De Moivre's theorem and its application

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De-Moivre’s theorem

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Let \( z=r \cos \theta +i\sin \theta \) be a complex number then \( z^n =r^n (\cos n\theta +i\sin n\theta )\) where n is a positive integer.
Proof

1. Case 1: n=1
   Then
   \( z =r (\cos \theta +i\sin \theta )\)
   or \( z^1 =r^1 (\cos 1.\theta +i\sin 1.\theta )\)
   So,
   or \( z^n =r^n (\cos n\theta +i\sin n\theta )\) when n=1
   
2. Case 2: n=2
   \( z^2 =z.z\)
   or \( z^2 = r (\cos \theta +i\sin \theta ) \times r (\cos \theta +i\sin \theta ) \)
   or \( z^2 = r^2 (\cos \theta +i\sin \theta )(\cos \theta +i\sin \theta )\)
   or \( z^2 = r^2 [\cos \theta (\cos \theta +i\sin \theta )+i\sin \theta (\cos \theta +i\sin \theta )\)
   or \( z^2 = r^2 [\cos \theta \cos \theta +i\cos \theta\sin \theta +i\sin \theta\cos \theta -\cos \theta\sin \theta ]\)
   or \( z^2 = r^2 [(\cos \theta \cos \theta-\sin \theta \sin \theta ) +i(\cos \theta\sin \theta +\sin \theta\cos \theta) ]\)
   or \( z^2 = r^2 [\cos (\theta +\theta) +i \sin (\theta+ \theta)]\)
   or \( z^2 = r^2 [\cos 2\theta +i \sin 2\theta]\)
   So,
   or \( z^n =r^n (\cos n\theta +i\sin n\theta )\) when n=2
   
3. Case 3: We assume the same formula is true for n = k, so we have
   \( (\cos\theta + i\sin\theta)^k = r^k(\cos(k\theta) + i\sin(k\theta))\)
   So,
   or \( z^n =r^n (\cos n\theta +i\sin n\theta )\) when n=k
   
4. Case 4: Now, we prove for n = k + 1,
   \( [r(\cos\theta + i\sin\theta)]^{k + 1} = r^k(\cos\theta + i\sin\theta)^k r (\cos\theta + i\sin\theta)\)
   or \( [r(\cos\theta + i\sin\theta)]^{k + 1} = r^k(\cos(k\theta) + i\sin(k\theta)) r(\cos\theta + i\sin\theta)\)
   or \( [r(\cos\theta + i\sin\theta)]^{k + 1} = r^{k+1}[(\cos(k\theta) \cos\theta-\sin(k\theta)\sin\theta )+i (\cos\theta\sin(k\theta) + \sin\theta\cos(k\theta))]\)
   or \( [r(\cos\theta + i\sin\theta)]^{k + 1} =r^{k+1}[\cos(k\theta+\theta)+i \sin(k\theta+\theta )]\)
   or \( [r(\cos\theta + i\sin\theta)]^{k + 1} =r^{k+1}[\cos(k+1)\theta+i \sin(k+1)\theta]\)
   So,
   or \( z^n =r^n (\cos n\theta +i\sin n\theta )\) when n=k+1
   
5. Using case 1-case 4, for any number \( n \in Z\) , we have
   \( [r(\cos\theta + i\sin\theta)]^n =r^n[\cos(n\theta)+i \sin(n\theta)]\)

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Exercise 1

1. Evaluate
   1. \( [2 (\cos 15^o+i \sin 15^o)]^6\)
      

      Answer 👉 Click Here

      Solution
      Given complex number is
      \( z=r(\cos \theta +i \sin \theta ) \)
      \( z=2(\cos 15^\circ + i \sin 15^\circ) \)
      

      Thus,

      \( z^6\)
      \( r^6 (\cos 6 \theta +i \sin 6 \theta ) \)
      \( 2^6 (\cos 6 \cdot 15^\circ +i \sin 6 \cdot 15^\circ ) \)
      \( 64 (\cos 90^\circ +i \sin 90^\circ ) \)
      \( 64 \left( 0 + i \cdot 1 \right) \)
      \(64i \)

      ---

   2. \( [3 (\cos 120^o+i \sin 120^o)]^3\)
      

      Answer 👉 Click Here

      Solution
      Given complex number is
      \( z=r(\cos \theta +i \sin \theta ) \)
      \( z=3(\cos 120^\circ + i \sin 120^\circ) \)
      

      Thus,

      \( z^3\)
      \(r^3 (\cos 3 \theta +i \sin 3 \theta ) \)
      \( 3^3 (\cos 3 \cdot 120^\circ +i \sin 3 \cdot 120^\circ ) \)
      \( 27 (\cos 360^\circ +i \sin 360^\circ ) \)
      \( 27 (\cos 0^\circ +i \sin 0^\circ ) \)
      \( 27 (1 + i \cdot 0) \)
      \(27 \)

      ---

   3. \( [ (\cos 18^o+i \sin 18^o)]^5\)
      

      Answer 👉 Click Here

      Solution
      Given complex number is
      \( z=r(\cos \theta +i \sin \theta ) \)
      \( z=\cos 18^\circ + i \sin 18^\circ \)
      

      Thus,

      \( z^5 \)
      \( r^5 (\cos 5 \theta +i \sin 5 \theta ) \)
      \( 1^5 (\cos 5 \cdot 18^\circ + i \sin 5 \cdot 18^\circ ) \)
      \( \cos 90^\circ + i \sin 90^\circ \)
      \( 0 + i \cdot 1 \)
      \( i \)

      ---

   4. \( [ (\cos 9^o+i \sin 9^o)]^{40}\)
      

      Answer 👉 Click Here

      Solution
      Given complex number is
      \( z=r(\cos \theta +i \sin \theta ) \)
      \( z=\cos 9^\circ + i \sin 9^\circ \)
      

      Thus,

      \( z^{40} \)
      \( r^{40} (\cos 40 \theta +i \sin 40 \theta ) \)
      \(1^{40} (\cos 40 \cdot 9^\circ + i \sin 40 \cdot 9^\circ ) \)
      \( \cos 360^\circ + i \sin 360^\circ \)
      \( \cos 0^\circ + i \sin 0^\circ \)
      \( 1 + i \cdot 0 \)
      \( 1 \)

      ---

   5. \( (1+i)^6\)
      Solution
      

      Answer 👉 Click Here

      Solution
      

      Given Complex number is
      \( z=1+i \)
      \( z=1+1i\)
      Comparing with x+iy, we get \( x=1,y=1\)
      So,
      \(r= \sqrt{(𝑥)^2+(𝑦)^2}=\sqrt{(1)^2+(1)^2}=\sqrt{2} \)
      \(\theta= \tan^{-1} \frac{y}{x} \tan^{-1} \frac{1}{1}= \tan^{-1} (1)= \frac{\pi}{4}\)
      So, we have
      \(r= \sqrt{2}, \theta= \frac{\pi}{4}\)
      Now,
      \( z=r(\cos \theta +i \sin \theta ) \)
      \( 1+i=\sqrt{2} (\cos \frac{\pi}{4}+i \sin \frac{\pi}{4} ) \)
      Thus,
      \( z^6 \)
      \( r^6(\cos 6 \theta +i \sin 6 \theta ) \)
      \( \sqrt{2}^6(\cos \frac{6\pi}{4} +i \sin \frac{6\pi}{4} ) \)
      \(8(\cos \frac{3\pi}{2} +i \sin \frac{3\pi}{2} ) \)
      \( 8[0 +i (-1) ] \)
      \( -8i \)

      ---

   6. \( (1+i)^{20}\)
      

      Answer 👉 Click Here

      Solution
      Given Complex number is
      \( z=1+i \)
      \( z=1+1i \)
      

      Comparing with x+iy, we get

      \( x=1, y=1 \)
      

      So,

      \( r= \sqrt{(𝑥)^2+(𝑦)^2}=\sqrt{(1)^2+(1)^2}=\sqrt{2} \)
      \(\theta= \tan^{-1} \frac{y}{x} = \tan^{-1} \frac{1}{1}= \tan^{-1} (1)= \frac{\pi}{4} \)
      

      So, we have

      \( r= \sqrt{2}, \theta= \frac{\pi}{4} \)
      

      Now,

      \( z=r(\cos \theta +i \sin \theta ) \)
      \( 1+i=\sqrt{2} (\cos \frac{\pi}{4} + i \sin \frac{\pi}{4} ) \)
      

      Thus,

      \( z^{20} \)
      \( r^{20} (\cos 20 \theta +i \sin 20 \theta ) \)
      \( \sqrt{2}^{20} (\cos 20 \cdot \frac{\pi}{4} +i \sin 20 \cdot \frac{\pi}{4} ) \)
      \( 2^{10} (\cos 5\pi +i \sin 5\pi ) \)
      \( 1024 (\cos 5\pi +i \sin 5\pi ) \)
      \( 1024 (\cos \pi +i \sin \pi ) \)
      \(1024 (-1 +i \cdot 0) \)
      \( -1024 \)

      ---

   7. \( (-1+i)^{14}\)
      

      Answer 👉 Click Here

      Solution
      Given Complex number is
      \( z=-1+i \)
      \( z=-1+1i \)
      

      Comparing with x+iy, we get

      \( x=-1, y=1 \)
      

      So,

      \( r= \sqrt{(𝑥)^2+(𝑦)^2}=\sqrt{(-1)^2+(1)^2}=\sqrt{2} \)
      \(\theta= \tan^{-1} \frac{y}{x} = \tan^{-1} \frac{1}{-1}= \tan^{-1} (-1)= 135^o \)
      

      So, we have

      \( r= \sqrt{2}, \theta= 135^o \)
      

      Now,

      \( z=r(\cos \theta +i \sin \theta ) \)
      \( -1+i=\sqrt{2} (\cos 135^o+ i \sin 135^o) \)
      

      Thus,

      \( z^{14}\)
      \(r^{14} (\cos 14 \theta +i \sin 14 \theta ) \)
      \( \sqrt{2}^{14} (\cos 14 \cdot (135^o) +i \sin 14 \cdot (135^o) ) \)
      \(\sqrt{2}^{14} [\cos (90)+i \sin (90)] \)
      \(128 [0 +i (1)] \)
      \(128i \)

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   8. \( \left ( \frac{1}{2} , \frac{\sqrt{3}}{2} i \right )^7 \)
      

      Answer 👉 Click Here

      Solution
      Given Complex number is
      \( z=\frac{1}{2} + \frac{\sqrt{3}}{2} i \)
      

      Comparing with x+iy, we get

      \( x=\frac{1}{2}, y=\frac{\sqrt{3}}{2} \)
      

      So,

      \( r= \sqrt{(𝑥)^2+(𝑦)^2}=\sqrt{\left(\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}=\sqrt{\frac{1}{4}+\frac{3}{4}}=\sqrt{1}=1 \)
      \(\theta= \tan^{-1} \frac{y}{x} = \tan^{-1} \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}= \tan^{-1} (\sqrt{3}) = 60^\circ \)
      

      So, we have

      \( r= 1, \theta= 60^\circ \)
      

      Now,

      \( z=r(\cos \theta +i \sin \theta ) \)
      \( \frac{1}{2} + \frac{\sqrt{3}}{2} i = 1 (\cos 60^\circ + i \sin 60^\circ) \)
      

      Thus,

      \( z^7\)
      \(r^7 (\cos 7 \theta +i \sin 7 \theta ) \)
      \( 1^7 (\cos 7 \cdot 60^\circ +i \sin 7 \cdot 60^\circ ) \)
      \( \cos 420^\circ +i \sin 420^\circ \)
      \( \cos 60^\circ +i \sin 60^\circ \)
      \( \frac{1}{2} + \frac{\sqrt{3}}{2} i \)

      ---

   9. \( (1-\sqrt{3}i)^6\)
      

      Answer 👉 Click Here

      Solution
      Given Complex number is
      \( z=1-\sqrt{3}i \)
      

      Comparing with x+iy, we get

      \( x=1, y=-\sqrt{3} \)
      

      So,

      \( r= \sqrt{(𝑥)^2+(𝑦)^2}=\sqrt{(1)^2+(-\sqrt{3})^2}=\sqrt{1+3}=\sqrt{4}=2 \)
      \(\theta= \tan^{-1} \frac{y}{x} = \tan^{-1} \frac{-\sqrt{3}}{1}= \tan^{-1} (-\sqrt{3}) = 300^o \)
      

      So, we have

      \( r= 2, \theta= 300^o \)
      

      Now,

      \( z=r(\cos \theta +i \sin \theta ) \)
      \( 1-\sqrt{3}i = 2 (\cos (300^o) + i \sin (300^o)) \)
      

      Thus,

      \( z^6\)
      \( r^6 (\cos 6 \theta +i \sin 6 \theta ) \)
      \( 2^6 (\cos 6 \cdot (300^o) +i \sin 6 \cdot (300^o) ) \)
      \( 64 (\cos 0^o +i \sin 0^o ) \)
      \( 64 (1 +i \cdot 0) \)
      \( 64 \)

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   10. \( i^2\)
       

       Answer 👉 Click Here

       Solution
       Given Complex number is
       \( z=i \)
       \( z=0+1i \)
       

       Comparing with x+iy, we get

       \( x=0, y=1 \)
       

       So,

       \( r= \sqrt{(𝑥)^2+(𝑦)^2}=\sqrt{(0)^2+(1)^2}=\sqrt{0+1}=\sqrt{1}=1 \)
       \(\theta= \tan^{-1} \frac{y}{x} = \tan^{-1} \frac{1}{0} = \frac{\pi}{2} \)
       

       So, we have

       \( r= 1, \theta= \frac{\pi}{2} \)
       

       Now,

       \( z=r(\cos \theta +i \sin \theta ) \)
       \( i = 1 (\cos \frac{\pi}{2} + i \sin \frac{\pi}{2}) \)
       

       Thus,

       \( z^2\)
       \(r^2 (\cos 2 \theta +i \sin 2 \theta ) \)
       \( 1^2 (\cos 2 \cdot \frac{\pi}{2} +i \sin 2 \cdot \frac{\pi}{2} ) \)
       \( \cos \pi +i \sin \pi \)
       \( -1 +i \cdot 0 \)
       \( -1 \)

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2. Show that \(z^3=1\) where \( z=\left ( -\frac{1}{2} , \frac{\sqrt{3}}{2} i \right ) \)
   

   Answer 👉 Click Here

   Solution
   Given Complex number is
   \( z=-\frac{1}{2} + \frac{\sqrt{3}}{2} i \)
   

   Comparing with x+iy, we get

   \( x=-\frac{1}{2}, y=\frac{\sqrt{3}}{2} \)
   

   So,

   \( r= \sqrt{(𝑥)^2+(𝑦)^2}=\sqrt{\left(-\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}=\sqrt{\frac{1}{4}+\frac{3}{4}}=\sqrt{1}=1 \)
   \(\theta= \tan^{-1} \frac{y}{x} = \tan^{-1} \frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}}= \tan^{-1} (-\sqrt{3}) = 120^\circ \)
   

   So, we have

   \( r= 1, \theta= 120^\circ \)
   

   Now,

   \( z=r(\cos \theta +i \sin \theta ) \)
   \( -\frac{1}{2} + \frac{\sqrt{3}}{2} i = 1 (\cos 120^\circ + i \sin 120^\circ) \)
   

   Thus,

   \( z^3 \)
   \(r^3 (\cos 3 \theta +i \sin 3 \theta ) \)
   \(1^3 (\cos 3 \cdot 120^\circ +i \sin 3 \cdot 120^\circ ) \)
   \( \cos 360^\circ +i \sin 360^\circ \)
   \( \cos 0^\circ +i \sin 0^\circ \)
   \( 1 +i \cdot 0 \)
   \( 1 \)
   This completes the solution

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nth root of Complex number

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If \( Z=r(\cos \theta +i\sin \theta )\) be a complex number then the nth root of z is
\( \sqrt[n]{Z}=\sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
Proof
Given that Z is a complex number. Also let, nth root of Z is W such that \( W=R(\cos \phi +i\sin \phi )\)
Now we have
\( \sqrt[n]{Z}=W\)
or \( W^n=Z\)
or \( [R(\cos \phi +i\sin \phi )]^n=r(\cos \theta +i\sin \theta )\)
or \( R^n(\cos (n\phi) +i\sin (n\phi) )=r(\cos \theta +i\sin \theta )\)
Equating real and Imaginary parts, we get
\( R^n=r\) and \( \cos (n\phi)= \cos \theta\) and \( \sin (n\phi) =\sin \theta \)
or \( R=\sqrt[n]{r}\) and \( n\phi= \theta +2k \pi \)
or \( R=\sqrt[n]{r}\) and \( \phi= \frac{(\theta +2k\pi )}{n}\)
Thus, nth root of \( Z=r(\cos \theta +i\sin \theta )\) is
\( W=R(\cos \phi +i\sin \phi )\)
or \( W=\sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)

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Exercise 2

1. Find the square roots of
   1. \(i\)

      Answer 👉 Click Here

      Given Complex number is
      \( z=i \)
      \( z=0+1i\)
      Comparing with x+iy, we get
      \( x=0,y=1\)
      So,
      \(r= \sqrt{(𝑥)^2+(𝑦)^2}=\sqrt{(0)^2+(1)^2}=1 \)
      \(\theta= \tan^{-1} \frac{y}{x} \tan^{-1} \frac{1}{0}= 90^o\)
      So, we have
      \(r= 1, \theta= 90^o\)
      

      Thus
      \( i=\cos 90^o +i\sin 90^o \)
      We know that, the nth roots of complex number Z are
      \( W=\sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \) for \(k=1,2,3..\)
      So, the two roots of z=i are

      1. \( w_1[i=0] \) taking k=0 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
         
         \(\cos \frac{90+0}{2} +i\sin \frac{90+0}{2}\)
         \(\cos 45+i\sin 45\) \( \frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}\)
      2. \( w_2[i=1]\) taking k=1 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
         
         \(\cos \frac{90+360 }{2} +i\sin \frac{90+360}{2}\)
         \(\cos 225+i\sin 225\)
         \(\frac{1}{\sqrt{2}}-i\frac{1}{\sqrt{2}}\)
   2. \(4+4\sqrt{3} i\)
      

      Answer 👉 Click Here

      Given Complex number is
      \( z = 4 + 4\sqrt{3}i \)
      Comparing with x + iy, we get
      \( x = 4, y = 4\sqrt{3} \)
      So,
      \( r = \sqrt{x^2 + y^2} = \sqrt{4^2 + (4\sqrt{3})^2} = \sqrt{16 + 48} = \sqrt{64} = 8 \)
      \( \theta = \tan^{-1} \frac{y}{x} = \tan^{-1} \frac{4\sqrt{3}}{4} = \tan^{-1} \sqrt{3} = 60^o \)
      So, we have
      \( r = 8, \theta = 60^o \)
      

      Thus
      \( 4 + 4\sqrt{3}i = 8(\cos 60^o + i\sin 60^o) \)
      We know that, the nth roots of complex number Z are
      \( W=\sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \) for \(k=1,2,3..\)
      So, two roots of \( z = 4 + 4\sqrt{3}i \) are

      1. \( w_1 [k=0]\) taking k=0 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
         
         \( \sqrt{8}(\cos \frac{60^o + 0}{2} + i\sin \frac{60^o + 0}{2}) \)
         \( \sqrt{8}(\cos 30^o + i\sin 30^o) \)
         \( 2\sqrt{2}(\frac{\sqrt{3}}{2} + i\frac{1}{2})\)
         \( \sqrt{6} + i\sqrt{2} \)
      2. \( w_2 [k=1]\) taking k=1 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
         
         \(\sqrt{8}(\cos \frac{60^o + 360^o}{2} + i\sin \frac{60^o + 360^o}{2}) \)
         \( \sqrt{8}(\cos 210^o + i\sin 210^o) \)
         \( 2\sqrt{2}(-\frac{\sqrt{3}}{2} - i\frac{1}{2}) \)
         \( -\sqrt{6} - i\sqrt{2} \)
   3. \(-1+\sqrt{3} i \)
      

      Answer 👉 Click Here

      Given Complex number is
      \( z = -1 + \sqrt{3}i \)
      Comparing with x + iy, we get
      \( x = -1, y = \sqrt{3} \)
      So,
      \( r = \sqrt{x^2 + y^2} = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2 \)
      \( \theta = \tan^{-1} \frac{y}{x} = \tan^{-1} \frac{\sqrt{3}}{-1} = \tan^{-1}(-\sqrt{3}) = 120^o \)
      So, we have
      \( r = 2, \theta =120^o \)
      

      Thus
      \( -1 + \sqrt{3}i = 2(\cos (120^o) + i\sin (120^o)) \)
      We know that, the nth roots of complex number Z are
      \( W=\sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \) for \(k=1,2,3..\)
      So, two roots of \( z = -1 + \sqrt{3}i \) are

      1. \( w_1 \) taking k=0 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
         
         \( \sqrt{2}(\cos \frac{120^o + 0}{2} + i\sin \frac{120^o + 0}{2}) \)
         \( \sqrt{2}(\cos 60^o + i\sin 60^o)\)
         \( \sqrt{2}(\frac{1}{2} + i\frac{\sqrt{3}}{2}) \)
         \( \frac{1}{\sqrt{2}}(1 + \sqrt{3}i ) \)
      2. \( w_2 \) taking k=1 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
         
         \( \sqrt{2}(\cos \frac{120^o + 360^o}{2} + i\sin \frac{120^o + 360^o}{2}) \)
         \( \sqrt{2}(\cos 240^o + i\sin 240^o) \)
         \( \sqrt{2}(-\frac{1}{2} + i\frac{\sqrt{3}}{2}) \)
         \( \frac{1}{\sqrt{2}}(-1 + \sqrt{3}i ) \)
   4. \(-2-2\sqrt{3} i \)
      

      Answer 👉 Click Here

      Given Complex number is
      \(z=-2-2\sqrt{3} i \)
      Comparing with x + iy, we get
      \( x = -2, y = -2\sqrt{3} \)
      So,
      \( r = \sqrt{x^2 + y^2} = \sqrt{(-2)^2 + (-2\sqrt{3})^2} = \sqrt{16} = 4 \)
      \( \theta = \tan^{-1} \frac{y}{x} = \tan^{-1} \frac{-2\sqrt{3}}{-2} = \tan^{-1} \sqrt{\sqrt{3}} = 240^o \)
      So, we have
      \( r = 4, \theta = 240^o \)
      

      Thus
      \( -2-2\sqrt{3} i = 4(\cos 240^o + i\sin 240^o) \)
      We know that, the nth roots of complex number Z are
      \( W=\sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \) for \(k=1,2,3..\)
      So, two roots of \(z=-2-2\sqrt{3} i \) are

      1. \( w_1 [k=0]\) taking k=0 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
         
         \( \sqrt{4}(\cos \frac{240^o + 0}{2} + i\sin \frac{240^o + 0}{2}) \)
         \( 2(\cos 120^o + i\sin 120^o) \)
         \( 2(-\frac{1}{2} + i\frac{\sqrt{3}}{2})\)
         \( -1 + \sqrt{3}i \)
      2. \( w_2 [k=1]\) taking k=1 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
         
         \( \sqrt{4}(\cos \frac{240^o + 360}{2} + i\sin \frac{240^o + 360}{2}) \)
         \( 2(\cos 300^o + i\sin 300^o) \)
         \( 2(\frac{1}{2} -i\frac{\sqrt{3}}{2})\)
         \( 1 - \sqrt{3}i \)
   5. \(2i\)
      

      Answer 👉 Click Here

      Given Complex number is
      \(z=0+2 i \)
      Comparing with x + iy, we get
      \( x = 0, y =2\)
      So,
      \( r = \sqrt{x^2 + y^2} = \sqrt{(0)^2 + (2)^2} = \sqrt{4} =2 \)
      \( \theta = \tan^{-1} \frac{y}{x} = \tan^{-1} \frac{2}{0} = 90^o \)
      So, we have
      \( r = 2, \theta = 90^o \)
      

      Thus
      \( 2 i = 2(\cos 90^o + i \sin 90^o) \)
      We know that, the nth roots of complex number Z are
      \( W=\sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \) for \(k=1,2,3..\)
      So, the two roots of \(z=2 i \) are

      1. \( w_1 [k=0]\) taking k=0 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
         
         \( \sqrt{2}(\cos \frac{90^o + 0}{2} + i \sin \frac{90^o + 0}{2}) \)
         \( \sqrt{2} (\cos 45^o + i\sin 45^o) \)
         \( \sqrt{2}( \frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}})\)
         \( 1 + 1i \)
      2. \( w_2 [k=1]\) taking k=1 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
         
         \( \sqrt{2}(\cos \frac{90^o + 360}{2} + i \sin \frac{90^o + 360}{2}) \)
         \( \sqrt{2} (\cos 225^o + i\sin 225^o) \)
         \( \sqrt{2}( - \frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}})\)
         \( - 1 + 1i \)
   6. \(-i \)
      

      Answer 👉 Click Here

      Given Complex number is
      \(z=0-1 i \)
      Comparing with x + iy, we get
      \( x = 0, y =-1\)
      So,
      \( r = \sqrt{x^2 + y^2} = \sqrt{(0)^2 + (-1)^2} = 1 \)
      \( \theta = \tan^{-1} \frac{y}{x} = \tan^{-1} \frac{-1}{0} = 270^o \)
      So, we have
      \( r = 1, \theta = 270^o \)
      

      Thus
      \( - i = 1(\cos 270^o + i \sin 27^o) \)
      We know that, the nth roots of complex number Z are
      \( W=\sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \) for \(k=1,2,3..\)
      So, the two roots of \(z=2 i \) are

      1. \( w_1 [k=0]\) taking k=0 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
         
         \( 1(\cos \frac{270^o + 0}{2} + i\sin \frac{270^o + 0}{2}) \)
         \( (\cos 135^o + i\sin 135^o) \)
         \( (- \frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}})\)
      2. \( w_2 [k=1]\) taking k=1 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
         
         \( 1(\cos \frac{270^o + 360}{2} + i\sin \frac{270^o + 360}{2}) \)
         \( (\cos 315^o + i\sin 315^o) \)
         \( (- \frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}})\)
2. Find the cube roots of
   1. \(8+6i\)
      

      Answer 👉 Click Here

      Given Complex number is
      \( z = 8 + 6i \)
      Comparing with x + iy, we get
      \( x = 8, y = 6 \)
      So,
      \( r = \sqrt{x^2 + y^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \)
      \( \theta = \tan^{-1} \frac{y}{x} = \tan^{-1} \frac{6}{8} = \tan^{-1} \frac{3}{4} \)
      So, we have
      \( r = 10, \theta = \tan^{-1} \frac{3}{4} = 36\)
      

      Thus
      \( 8 + 6i = 10(\cos 36+ i\sin 36) \)
      We know that, the nth roots of complex number Z are
      \( W=\sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \) for \(k=1,2,3..\)
      So, the three roots of \( z = 8 + 6i \) are

      1. \( w_1\) taking k=0 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
         
         \( \sqrt[3]{10}(\cos \frac{36+ 0}{3} + i\sin \frac{36+ 0}{3}) \)
         \( \sqrt[3]{10}(\cos 12 + i\sin 12) \)
      2. \( w_2\) taking k=1 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
         
         \( \sqrt[3]{10}(\cos \frac{36+ 360}{3} + i\sin \frac{36+ 360}{3}) \)
         \( \sqrt[3]{10}(\cos 132 + i\sin 132) \)
      3. \( w_3\) taking k=2 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
         
         \( \sqrt[3]{10}(\cos \frac{36+ 720}{3} + i\sin \frac{36+720}{3}) \)
         \( \sqrt[3]{10}(\cos 252+ i\sin 252) \)
   2. \(-1\)

      Answer 👉 Click Here

      Given Complex number is
      \( z = -1+0i \)
      Comparing with x + iy, we get
      \( x =-1, y = 0 \)
      So,
      \( r = \sqrt{x^2 + y^2} = \sqrt{(-1)^2 + (0)^2} = 1\)
      \( \theta = \tan^{-1} \frac{y}{x} = \tan^{-1} \frac{0}{-1} = \tan^{-1} 0= 180^o\)
      So, we have
      \( r = 1, \theta = 180\)
      

      Thus
      \( -1 = 1(\cos 180+ i\sin 180) \)
      We know that, the nth roots of complex number Z are
      \( W=\sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \) for \(k=1,2,3..\)
      So, the three roots of \( z = -1 \) are

      1. \( w_1\) taking k=0 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
         
         \( (\cos \frac{180+ 0}{3} + i\sin \frac{180+ 0}{3}) \)
         \( (\cos 60 + i\sin 60) \)
         \( (\frac{\sqrt{3}}{2} + \frac{1}{2} i) \)
      2. \( w_2\) taking k=1 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
         
         \( (\cos \frac{180+ 360}{3} + i\sin \frac{180+ 360}{3}) \)
         \( (\cos 180 + i\sin 180) \)
         \( (0 -1 i) \)
      3. \( w_2\) taking k=2 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
         
         \( (\cos \frac{180+ 720}{3} + i\sin \frac{180+ 720}{3}) \)
         \( (\cos 300 + i\sin 300) \)
         \( (-\frac{\sqrt{3}}{2} - \frac{1}{2} i) \)
   3. \(8i\)

      Answer 👉 Click Here

      Given Complex number is
      \( z = 0+8i \)
      Comparing with x + iy, we get
      \( x =0, y =8 \)
      So,
      \( r = \sqrt{x^2 + y^2} = \sqrt{(0)^2 + (8)^2} = 8\)
      \( \theta = \tan^{-1} \frac{y}{x} = \tan^{-1} \frac{8}{0} = 90^o\)
      So, we have
      \( r =8, \theta = 90\)
      

      Thus
      \( 8i = 1(\cos 180+ i\sin 180) \)
      We know that, the nth roots of complex number Z are
      \( W=\sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \) for \(k=1,2,3..\)
      So, the three roots of \( z = -1 \) are

      1. \( w_1\) taking k=0 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
         
         \( 2 \sqrt{2} (\cos \frac{90+ 0}{3} + i\sin \frac{90+ 0}{3}) \)
         \(2 \sqrt{2} (\cos 30 + i\sin 30) \)
         \( 2 \sqrt{2} (\frac{\sqrt{3}}{2} + \frac{1}{2} i) \)
         \( \sqrt{2} (\sqrt{3}+ i) \)
      2. \( w_2\) taking k=1 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
         
         \( 2 \sqrt{2} (\cos \frac{90+ 360}{3} + i\sin \frac{90+ 360}{3}) \)
         \(2 \sqrt{2} (\cos 150 + i\sin 150) \)
         \( 2 \sqrt{2} (-\frac{\sqrt{3}}{2} + \frac{1}{2} i) \)
         \( \sqrt{2} (-\sqrt{3}+ i) \)
      3. \( w_3\) taking k=2 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
         
         \( 2 \sqrt{2} (\cos \frac{90+ 720}{3} + i\sin \frac{90+ 720}{3}) \)
         \(2 \sqrt{2} (\cos 270 + i\sin 270) \)
         \( 2 \sqrt{2} (0 - i) \)
         \( 2 \sqrt{2} i \)
3. Solve the following
   1. \(z^4=1\)
      

      Answer 👉 Click Here

      Given Complex number is
      \( z^4 = 1\)
      We know that every polynomial of degree n has n roots, so, z has four roots.
      Comparing 1 with x + iy, we get
      \( 1=1+0.i\)
      \( x =1, y =0 \)
      So,
      \( r = \sqrt{x^2 + y^2} = \sqrt{(1)^2 + (0)^2} = 1\)
      \( \theta = \tan^{-1} \frac{y}{x} = \tan^{-1} \frac{0}{1} = 0^o\)
      So, we have
      \( r =1, \theta = 0\)
      

      Thus
      \( 1 = 1(\cos 0+ i\sin 0) \)
      We know that, the nth roots of complex number Z are
      \( W=\sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \) for \(k=1,2,3..\)
      So, the four roots of \( z^4=1 \) are

      1. \( w_1\) taking k=0 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
         \( 1 (\cos \frac{0+ 0}{4} + i\sin \frac{0+ 0}{4}) \)
         \((\cos 0 + i\sin 0) \)
         \( 1 + 0. i \)
         \( 1 \)
      2. \( w_2\)taking k=1 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
         \( 1 (\cos \frac{360+ 0}{4} + i\sin \frac{360+ 0}{4}) \)
         \((\cos 90 + i\sin 90) \)
         \( 0 + 1. i \)
         \( i \)
      3. \( w_3\)taking k=2 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
         \( 1 (\cos \frac{720+ 0}{4} + i\sin \frac{720+ 0}{4}) \)
         \((\cos 180 + i\sin 100) \)
         \( -1 + 0. i \)
         \( -1 \)
      4. \( w_4\)taking k=3 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
         \( 1 (\cos \frac{1080+ 0}{4} + i\sin \frac{1080+ 0}{4}) \)
         \((\cos 270 + i\sin 270) \)
         \( 0 - 1. i \)
         \( -i \)
   2. \(z^6=1\)
      

      Answer 👉 Click Here

      Given Complex number is
      \( z^6 = 1\)
      We know that every polynomial of degree n has n roots, so, z has six roots.
      Comparing 1 with x + iy, we get
      \( 1=1+0.i\)
      \( x =1, y =0 \)
      So,
      \( r = \sqrt{x^2 + y^2} = \sqrt{(1)^2 + (0)^2} = 1\)
      \( \theta = \tan^{-1} \frac{y}{x} = \tan^{-1} \frac{0}{1} = 0^o\)
      So, we have
      \( r =1, \theta = 0\)
      

      Thus
      \( 1 = 1(\cos 0+ i\sin 0) \)
      We know that, the nth roots of complex number Z are
      \( W=\sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \) for \(k=1,2,3..\)
      So, the six roots of \( z^6=1 \) are

      1. \( w_1\), taking k=0 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
         \( 1 (\cos \frac{0+ 0}{6} + i\sin \frac{0+ 0}{6}) \)
         \((\cos 0 + i\sin 0) \)
         \( 1 + 0. i \)
         \( 1 \)
      2. \( w_2\) taking k=1 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
         \( 1 (\cos \frac{360+ 0}{6} + i\sin \frac{360+ 0}{6}) \)
         \((\cos 60 + i\sin 60) \)
         \( \frac{1}{2} + \frac{\sqrt{3}}{2}. i \)
         \( \frac{1+ \sqrt{3} i}{2} \)
      3. \( w_3\) taking k=2 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
         \( 1 (\cos \frac{720+ 0}{6} + i\sin \frac{720+ 0}{6}) \)
         \((\cos 120 + i\sin 120) \)
         \( -\frac{1}{2} + \frac{\sqrt{3}}{2}. i \)
         \( \frac{-1+ \sqrt{3} i}{2} \)
      4. \( w_4\) taking k=3 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
         \( 1 (\cos \frac{1080+ 0}{6} + i\sin \frac{1080+ 0}{6}) \)
         \((\cos 180 + i\sin 180) \)
         \( -1 + 0. i \)
         \( -1 \)
      5. \( w_5\) taking k=4 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
         \( 1 (\cos \frac{1440+ 0}{6} + i\sin \frac{1440+ 0}{6}) \)
         \((\cos 240 + i\sin 240) \)
         \( -\frac{1}{2} - \frac{\sqrt{3}}{2}. i \)
         \( \frac{-1- \sqrt{3} i}{2} \)
      6. \( w_6\) taking k=5 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
         \( 1 (\cos \frac{1800+ 0}{6} + i\sin \frac{1800+ 0}{6}) \)
         \((\cos 300 + i\sin 300) \)
         \( \frac{1}{2} - \frac{\sqrt{3}}{2}. i \)
         \( \frac{1- \sqrt{3} i}{2} \)
   3. \(z^4+1=0\)
      

      Answer 👉 Click Here

      Given Complex number is
      \( z^4 = -1\)
      We know that every polynomial of degree n has n roots, so, z has four roots.
      Comparing 1 with x + iy, we get
      \( 1=-1+0.i\)
      \( x =-1, y =0 \)
      So,
      \( r = \sqrt{x^2 + y^2} = \sqrt{(-1)^2 + (0)^2} = 1\)
      \( \theta = \tan^{-1} \frac{y}{x} = \tan^{-1} \frac{0}{-1} = 180^o\)
      So, we have
      \( r =1, \theta = 180\)
      

      Thus
      \( -1 = 1(\cos 180+ i\sin 180) \)
      We know that, the nth roots of complex number Z are
      \( W=\sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \) for \(k=1,2,3..\)
      So, the four roots of \( z^4=-1 \) are

      1. \( w_1\) taking k=0 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
         \( 1 (\cos \frac{0+ 180}{4} + i\sin \frac{0+ 180}{4}) \)
         \((\cos 45 + i\sin 45) \)
         \( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}. i \)
         \( \frac{1+i}{\sqrt{2}} \)
      2. \( w_2\) taking k=1 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
         \( 1 (\cos \frac{360+ 180}{4} + i\sin \frac{360+ 180}{4}) \)
         \((\cos 135 + i\sin 135) \)
         \( -\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}. i \)
         \( \frac{-1+i}{\sqrt{2}} \)
      3. \( w_3\) taking k=2 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
         \( 1 (\cos \frac{720+ 180}{4} + i\sin \frac{720+ 180}{4}) \)
         \((\cos 225 + i\sin 225) \)
         \( -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}. i \)
         \( \frac{-1-i}{\sqrt{2}} \)
      4. \( w_4\) taking k=3 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
         \( 1 (\cos \frac{1080+ 180}{4} + i\sin \frac{1080+ 180}{4}) \)
         \((\cos 315 + i\sin 315) \)
         \( \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}. i \)
         \( \frac{1-i}{\sqrt{2}} \)
   4. \(z^3=8i\)
      

      Answer 👉 Click Here

      Given Complex number is
      \(z^3=8i\)
      We know that every polynomial of degree n has n roots, so, z has three roots.
      Comparing 1 with x + iy, we get
      \( 1=0+8.i\)
      \( x =0, y =8 \)
      So,
      \( r = \sqrt{x^2 + y^2} = \sqrt{(0)^2 + (8)^2} = 8\)
      \( \theta = \tan^{-1} \frac{y}{x} = \tan^{-1} \frac{8}{0} = 90^o\)
      So, we have
      \( r =8, \theta = 90\)
      

      Thus
      \( 8i = 8(\cos 90+ i\sin 90) \)
      We know that, the nth roots of complex number Z are
      \( W=\sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \) for \(k=1,2,3..\)
      So, the four roots of \( z^3=8i \) are

      1. \( w_1\) taking k=0 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
         \( \sqrt[3]{8} (\cos \frac{0+ 90}{3} + i\sin \frac{0+ 90}{3}) \)
         \( 2 (\cos 30 + i\sin 30) \)
         \( 2( \frac{\sqrt{3}}{2} + \frac{1}{2}. i) \)
         \( \sqrt{3}+i \)
      2. \( w_2\) taking k=1 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
         \( \sqrt[3]{8} (\cos \frac{360+ 90}{3} + i\sin \frac{360+ 90}{3}) \)
         \( 2 (\cos 150 + i\sin 150) \)
         \( 2( -\frac{\sqrt{3}}{2} + \frac{1}{2}. i) \)
         \( -\sqrt{3}+i \)
      3. \( w_3\) taking k=2 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
         \( \sqrt[3]{8} (\cos \frac{720+ 90}{3} + i\sin \frac{720+ 90}{3}) \)
         \( 2 (\cos 270 + i\sin 270) \)
         \( 2( 0 + 1. i) \)
         \(2i \)
4. Find the 4th roots of \( ( -\frac{1}{2},\frac{\sqrt{3}}{2}i ) \)
   

   Answer 👉 Click Here

   Given Complex number is
   \( z= -\frac{1}{2},\frac{\sqrt{3}}{2}i \)
   We know that every polynomial of degree n has n roots, so, z has four roots.
   Comparing 1 with x + iy, we get
   \( z= -\frac{1}{2},\frac{\sqrt{3}}{2}i \)
   \( x =-\frac{1}{2}, y =\frac{\sqrt{3}}{2} \)
   So,
   \( r = \sqrt{x^2 + y^2} = \sqrt{(-\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} = 1\)
   \( \theta = \tan^{-1} \frac{y}{x} = \tan^{-1} \frac{ \frac{\sqrt{3}}{2}}{-\frac{1}{2}} = 120^o\)
   So, we have
   \( r =1, \theta = 120\)
   

   Thus
   \( -\frac{1}{2},\frac{\sqrt{3}}{2}i = 1(\cos 120+ i\sin 120) \)
   We know that, the nth roots of complex number Z are
   \( W=\sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \) for \(k=1,2,3..\)
   So, the four roots of \( z \) are

   1. \( w_1\) taking k=0 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
      \( 1 (\cos \frac{0+ 120}{4} + i\sin \frac{0+ 120}{4}) \)
      \((\cos 30 + i\sin 30) \)
      \( \frac{\sqrt{3}}{2} + \frac{1}{2}. i \)
      \( \frac{\sqrt{3}+i}{2} \)
   2. \( w_2\) taking k=1 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
      \( 1 (\cos \frac{360+ 120}{4} + i\sin \frac{360+ 120}{4}) \)
      \((\cos 120 + i\sin 120) \)
      \( -\frac{1}{2} + \frac{\sqrt{3}}{2}. i \)
      \( \frac{-1+\sqrt{3}i}{2} \)
   3. \( w_3\) taking k=2 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
      \( 1 (\cos \frac{720+ 120}{4} + i\sin \frac{720+ 120}{4}) \)
      \((\cos 210 + i\sin 210) \)
      \( -\frac{\sqrt{3}}{2} - \frac{1}{2}. i \)
      \( \frac{-\sqrt{3}-i}{2} \)
   4. \( w_4\) taking k=3 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
      \( 1 (\cos \frac{1080+ 120}{4} + i\sin \frac{1080+ 120}{4}) \)
      \((\cos 300 + i\sin 300) \)
      \( -\frac{1}{2} + \frac{\sqrt{3}}{2}. i \)
      \( \frac{-1+\sqrt{3}i}{2} \)
5. Find the 10th roots of 1
   The figure shows 10th root of 1.
   Drag the value of k=0,1,2,...,9
6. In electrical engineering, a circuit has an impedance represented by the complex number. Z=8+6i ohms. The engineers need to design a component with an impedance that, when cubed, matches the original impedance.
   1. Calculate the magnitude ∣Z∣ and angle θ of the original impedance.
   2. Determine the cube root of the original impedance in polar form.
   3. Design a new component with an impedance Zn such that (Zn)^3 matches the original impedance.
   4. Express the new impedance in rectangular form and calculate its magnitude and angle.
7. If \( \bar{z}\) be the conjugate of a complex number \(z\), prove that \(Arg(\bar{z})=2 \pi- Arg(z)\)
8. If \(z=\cos \theta +i \sin \theta\), prove that \(z^n-\frac{1}{z^n} = 2 \sin n \theta i\)

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Properties of cube root of unity

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Exponential Form

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Euler's formula establishes the fundamental relationship between the trigonometric functions and the complex exponential function. It states that for any real number x:
\( e^{i \theta}=\cos \theta+i\sin \theta\)
where e is the base of the natural logarithm, i is the imaginary unit, and cos and sin are the trigonometric functions.

When x = π, Euler's formula evaluates to \( e^{i\pi}+1=0\), which is known as Euler's identity.

Given Euler’s exponential form,
\( e^{i\theta}=\cos \theta +i\sin \theta \)
Thus, complex number \( z=r( \cos \theta +i\sin \theta )\) is defined as
\( z=re^{i\theta}\)
The significance of exponential form of complex number is that we can easily compute conjugate and inverse.
For example,
\( \bar{z}=re^{- i\theta}\) and \( z^{-1} =\frac{1}{r} e^{- i\theta}\)

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Proof of the Formula

1. Function Method
   

   Consider the function f(θ) given by
   \( f(\theta )=\frac {\cos \theta +i\sin \theta }{e^{i\theta }}\)
   or\( f(\theta )= e^{- i\theta } ( \cos \theta +i\sin \theta) \)
   Differentiating gives by the product rule, we have
   \( f' (\theta )=0\)
   Thus, f(θ) is a constant.
   Since f(0) = 1, then f(θ) = 1 for all θ, and thus
   \( 1= e^{- i\theta } ( \cos \theta +i\sin \theta)\)
   or\( e^{ i\theta }= \cos \theta +i\sin \theta\)
   This completes the proof.

2. Series Method
   

   We know that
   \(e^x=\displaystyle \sum_{n=0}^\infty \frac{x^n}{n!} \)
   \(\cos x= 1- \frac{x^2}{2!}+ \frac{x^4}{4!}-\)
   \(\sin x= x-\frac{x^3}{3!} +\frac{x^5}{5!}-... \)
   Using power series expansion, we get
   \(e^{ i\theta }= \sum_{n=0}^\infty \frac{(i\theta)^n}{n!}\)
   or \(e^{ i\theta }= 1+ \frac{(i\theta)^1}{1!}+ \frac{(i\theta)^2}{2!} + \frac{(i\theta)^3}{3!}+ \frac{(i\theta)^4}{4!}+...\)
   or \(e^{ i\theta }= 1+ i\theta- \frac{\theta^2}{2!} -i \frac{\theta^3}{3!}+ \frac{\theta^4}{4!}+...\)
   or \(e^{ i\theta }= \left ( 1- \frac{\theta^2}{2!}+ \frac{\theta^4}{4!}-...\right ) +i \left (\theta-\frac{\theta^3}{3!} +... \right ) \)
   or \(e^{ i\theta } = \cos \theta +i\sin \theta\)
   This completes the proof.

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Tangent and Normal to a Circle

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Introduction to circle

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Circle is defined a locus of point whose
distance from a fixed point = constant

In this definition of circle,

• the constant distance is called radius.
• the fixed point is called center

Parts of Circle

1. Center: The fixed point from which the distance to circumference is constant is called center .
   O is Center

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2. Radius: A segment drawn from the center to the circumference is a called radius.
   OA is Radius

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3. Circumference: The locus of point which is equidistant from center is called circumference.
   The locus is circumference

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4. Diameter: A segment drawn through the center and ends to the circumference is called diameter.
   AB is diameter

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5. Chord: A segment that joins any two points on the circumference is called segment. Diameter is the longest chord.
   AB is a Chord

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6. Arc: A continuous piece of circumference is called an arc. There are two types of arc: Major (greater than half) and Minor (less than half). An arc half to the circle is called semi-circle.
   AB (clockwise) is majorr Arc, AB (anti-clockwise) is minor Arc

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7. Sector: A region enclosed by an arc with two radii and joining to the center is called sector. There are two types of sector: Major (greater than half) and Minor (less than half). A sector half to the circle is called semi-circle.
   AB(clockwise) is major Sector and AB(anti-clockwise) is minor Sector

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8. Segment: A region enclosed by an arc with a chord is called segment. There are two types of segment: Major (greater than half) and Minor (less than half). A segment half to the circle is called semi-circle.
   AB (anti-clockwise) is minor segment

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9. Tangent: A straight line that touches the circle exactly once but does not cut it, however extended, is called tangent.

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10. Secant: A straight line that cuts a circle at two points. It is extension of chord.

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11. Central Angle: An angle made at the center of a circle is called central angle.

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Equation of circle

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1. center at origin: \(x^2+y^2=r^2\)
   center= (0,0)
   radius= r
   
2. center at any point (h,k): \((x-h)^2+(y-k)^2=r^2\)
   center= (h,k)
   radius= r
   
3. general equation of circle : \(x^2+y^2+2gx+2fy+c=0\)
   center= (-g,-f)
   radius= \(\sqrt{g^2+f^2-c} \)
4. If two end points \(A=(x_1,y_1)\) and \(B=(x_2,y_2)\) of a diameter of a circle is given, then
   Method 1
   
   1. Find the distance between the endpoints \(A=(x_1,y_1)\) and \(B=(x_2,y_2)\) given by
      \(d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)
   2. Find the radius of circle given by
      \(r= \frac{1}{2} \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)
   3. Find the center of the circle given by
      \(h= \frac{x_1+x_2}{2}\)
      \(k= \frac{y_1+y_2}{2}\)
   4. Find the equation of the circle
      \((x-h)^2+(y-k)^2=r^2\)
      \((x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0\)

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   Method 2
   If two end points \(A=(x_1,y_1)\) and \(B=(x_2,y_2)\) of a diameter of a circle is given, then
   1. take any arbitrary point on the circle given by
      \(P=(x,y)\)
   2. Find slope of both lines at P
      \(m_1= \frac{y-y_1}{x-x_1} \)
      \(m_2= \frac{y-y_2}{x-x_2} \)
   3. Find the center of the circle given by
      \(m_1.m_2=-1\)
      \((x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0\)

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Partcular case of Circle

Let center equation of a circle be \((x-h)^2+(y-k)^2=r^2\) in which
center= (h,k)
radius= r
Then

1. Equation of the circle touching the x-axis
   \((x-h)^2+(y-k)^2=k^2\)
2. Equation of the circle touching the y-axis
   \((x-h)^2+(y-k)^2=h^2\)
3. Equation of the circle touching the both axis
   \((x-h)^2+(y-h)^2=h^2\)

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Tangent to a circle

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The tangent line to a circle is a straight line touching the circumference of the circle at only one point. A line is only a tangent if there is exactly one point of contact between the straight line and the circle.
Let C be a circle. Also let A and B are two points on C
when, point B is moved toward A then limiting form of secant AB, is called tangent to the circle C at A.

To find the equation of a tangent line, we first need to be able to find the gradient of a line, either by
\(m=\frac{raise}{run}\) or \(m=\frac{y^2-y^1}{x^2-x^1}\) or \(m=\frac{dy}{dx}\)
at the point of contact.

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Condition to be a Tangent to a circle

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1. Let, a circle is given by
   \(C:x^2+y^2=r^2\) (i)
   Also let a straight line is given by
   \(l: ax+by+c=0\) (ii)
   If line (ii) is tangent to the circle (i), then we can say that
   1. the line (ii) intersect the circle (i) exactly in one single point, called the Point of tangency(contact), say P.
   2. at the point of tangency P, the tangent (ii) is perpendicular to the radius of circle (i)
   3. at P, the ⊥ length from tangent line to (0,0) is radius.
   4. \( r=\pm \frac{Ax_1+By_1+C}{\sqrt{A^2+B^2}}\)
      \( r=\pm \frac{a.(0)+b.(0)+c}{\sqrt{a^2+b^2}}\)
      \( r^2 (a^2+b^2)=\pm c^2\)
   So
   \( r^2 (a^2+b^2)=\pm c^2\)
   is the required condition

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2. Let, a circle is given by
   \(C:x^2+y^2=r^2\) (i)
   Also let a straight line is given by
   \(l: y=mx+c\) (ii)
   If line (ii) is tangent to the circle (i), then we can say that
   1. the line (ii) intersect the circle (i) exactly in one single point, called the Point of tangency(contact), say P.
   2. at the point of tangency P, the tangent (ii) is perpendicular to the radius of circle (i)
   3. at P, the ⊥ length from tangent line to (0,0) is radius.
   4. \( r=\pm \frac{Ax_1+By_1+C}{\sqrt{A^2+B^2}}\)
      \( r=\pm \frac{m.(0)-(0)+c}{\sqrt{m^2+1^2}}\)
      \( r^2 (m^2+1)=\pm c^2\)
   So
   \( r^2 (m^2+1)=\pm c^2\)
   is the required condition

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3. Let, a circle is given by
   \(C:(x-h)^2+(y-k)^2=r^2\) (i)
   Also let a straight line is given by
   \(l: ax+by+c=0\) (ii)
   If line (ii) is tangent to the circle (i), then we can say that
   1. the line (ii) intersect the circle (i) exactly in one single point, called the Point of tangency(contact), say P.
   2. at the point of tangency P, the tangent (ii) is perpendicular to the radius of circle (i)
   3. at P, the ⊥ length from tangent line to (h,k) is radius.
   4. \( r=\pm \frac{Ax_1+By_1+C}{\sqrt{A^2+B^2}}\)
      \( r=\pm \frac{a(h)+b(k)+c}{\sqrt{a^2+b^2}}\)
      \( r^2 (a^2+b^2)=\pm (ah+bk+c)^2\)
   So
   \( r^2 (a^2+b^2)=\pm (ah+bk+c)^2\)
   is the required condition

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4. Let, a circle is given by
   \(C:(x-h)^2+(y-k)^2=r^2\) (i)
   Also let a straight line is given by
   \(l: y=mx+c\) (ii)
   If line (ii) is tangent to the circle (i), then we can say that
   1. the line (ii) intersect the circle (i) exactly in one single point, called the Point of tangency(contact), say P.
   2. at the point of tangency P, the tangent (ii) is perpendicular to the radius of circle (i)
   3. at P, the ⊥ length from tangent line to (h,k) is radius.
   4. \( r=\pm \frac{Ax_1+By_1+C}{\sqrt{A^2+B^2}}\)
      \( r=\pm \frac{m.(h)-(k)+c}{\sqrt{m^2+1^2}}\)
      \( r^2 (m^2+1)=\pm (mh-k+c)^2\)
   So
   \( r^2 (m^2+1)=\pm (mh-k+c)^2\)
   is the required condition

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5. Let, a circle is given by
   \(C:x^2+y^2+2gx+2fy+k=0\) (i)
   Also let a straight line is given by
   \(l: ax+by+c=0\) (ii)
   If line (ii) is tangent to the circle (i), then we can say that
   1. the line (ii) intersect the circle (i) exactly in one single point, called the Point of tangency(contact), say P.
   2. at the point of tangency P, the tangent (ii) is perpendicular to the radius of circle (i)
   3. at P, the ⊥ length from tangent line to (h,k) is radius, where (h,k)=(-g,-f)
   4. \( r=\pm \frac{Ax_1+By_1+C}{\sqrt{A^2+B^2}}\)
      \( \sqrt{f^2+g^2-k}=\pm \frac{a.(h)+b(k)+c}{\sqrt{a^2+b^2}}\)
      \( (f^2+g^2-k)^2 (a^2+b^2)=\pm (ah+bk+c)^2\)
   So
   \( (f^2+g^2-k)^2 (a^2+b^2)=\pm (ah+bk+c)^2\)
   is the required condition

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Equation of tangent line to the circle

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1. Equation of tangent line to the circle \(x^2+y^2=r^2\)
   Method 1

   Let \(C:x^2+y^2=r^2\) be a circle whose center is at O and let \(P(x_1,y_1)\) be a given point on C.
   Then
   \( x_1^2+y_1^2=r^2\)
   Also
   Equation of straight line passing through P is
   \( y-y_1=m(x-x_1)\)
   Now, the line OP has slope
   \(\frac{y_1}{x_1}\)
   So, the tangent line has slope
   \(m=-\frac{x_1}{y_1}\)
   Hence, equation of tangent line is
   \( y-y_1=m(x-x_1)\)
   or \( y-y_1=-\frac{x_1}{y_1} (x-x_1)\) Standard form
   or \( xx_1+yy_1=r^2\)
   This completes the proof

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   Method 2

   Let \(C:x^2+y^2=r^2\) be a circle whose center is at O and let \(P(x_1,y_1)\) be a given point on C.
   Then
   \( x_1^2+y_1^2=r^2\)
   Also
   Equation of straight line passing through P is
   \( y-y_1=m(x-x_1)\)
   Now, the slope of the tangent to the circle is
   \(m=\frac{dy}{dx}\)
   or \( m=-\frac{x_1}{y_1}\) at \(P(x_1,y_1)\)
   Hence, equation of tangent line is
   \( y-y_1=m(x-x_1)\)
   or \( y-y_1=-\frac{x_1}{y_1} (x-x_1)\) Standard form
   or \( xx_1+yy_1=r^2\)
   This completes the proof

2. Equation of tangent line to the circle \((x-h)^2+(y-k)^2=r^2\)
   

   Let \(C: (x-h)^2+(y-k)^2=r^2\)be a circle whose center is at A(h,k) and let \(P(x_1,y_1)\) be a given point on C.
   Then
   Equation of straight line passing through P is
   \( y-y_1=m(x-x_1)\)
   Now, the slope of the tangent to the circle is
   \(m=\frac{dy}{dx}\)
   or \( m=-\frac{x_1-h}{y_1-k}\) at \(P(x_1,y_1)\)
   Hence, equation of tangent line is
   \( y-y_1=m(x-x_1)\)
   or \( y-y_1=-\frac{x_1-h}{y_1-k} (x-x_1)\) Standard form
   or \( (x-x_1)(x_1-h)+(y-y_1)(y_1-k)=0\)
   This completes the proof

3. Equation of tangent line to the circle \(x^2+y^2+2gx+2fy+c=0\)
   

   Let \(C: x^2+y^2+2gx+2fy+c=0\) be a circle whose center and let \(P(x_1,y_1)\) be a given point on C.
   Then
   \(x_1^2+y_1^2+2gx_1+2fy_1+c=0\) (i)
   Also let \(Q(x_2,y_2)\) be a nearby point on C, then
   \( x_2^2+y_2^2+2gx_2+2fy_2+c=0\) (ii)
   Subtracting (i) from (ii), we get
   \( (x_2^2-x_1^2)+(y_2^2-y_1^2)+2g(x_2-x_1)+2f(y_2-y_1)=0\)
   or \( (x_2-x_1)(x_1+x_2+2g)+(y_2-y_1)(y_1+y_2+2f)=0\)
   or \( \frac{y_2-y_1}{x_2-x_1} =- \frac{x_1+x_2+2g}{y_1+y_2+2f} \)
   We know that, tangent is the limiting position of secant, so in the tangent line, the points \(P(x_1,y_1)\) and \(Q(x_2,y_2)\) coincides
   Now, the slope of tangent to the circle at \(P(x_1,y_1)\) is
   \(m=- \frac{x_1+x_2+2g}{y_1+y_2+2f} \)
   or \( m=-\frac{2x_1+2g}{2y_1+2f}\)
   or \( m=-\frac{x_1+g}{y_1+f} \)
   Hence, the equation of tangent line to the circle at \(P(x_1,y_1)\) is
   \( y-y_1=m(x-x_1)\)
   or \( y-y_1=-\frac{x_1+g}{y_1+f} (x-x_1)\) Standard form
   or \( xx_1+yy_1+gx+fy=x_1^2+y_1^2+gx_1+fy_1\)
   Adding \(x_1g+y_1f+c\) to both sides, we get
   \( xx_1+yy_1+gx+fy\) + \( x_1g+y_1f+c\) = \(x_1^2+y_1^2+gx_1+fy_1\) + \( x_1g+y_1f+c\)
   or \( xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=x_1^2+y_1^2+2gx_1+2fy_1+c\)
   or \( xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0\)
   This completes the proof

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Equation of normal line to the circle

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1. Equation of normal line to the circle \(x^2+y^2=r^2\)
   

   Let \(C:x^2+y^2=r^2\) be a circle whose center is at O and let \(P(x_1,y_1)\) be a given point on C.
   Then
   \( x_1^2+y_1^2=r^2\)
   Also
   Equation of straight line passing through P is
   \( y-y_1=m(x-x_1)\)
   Now, the line OP has slope
   \(\frac{y_1}{x_1}\)
   So, the normal line has slope
   \(m=\frac{y_1}{x_1}\)
   Hence, equation of normal line is
   \( y-y_1=m(x-x_1)\)
   or \( y-y_1=\frac{y_1}{x_1} (x-x_1)\) Standard form
   or \( xy_1-yx_1=0\)
   This completes the proof

2. Equation of normal line to the circle \((x-h)^2+(y-k)^2=r^2\)
   

   Let \(C: (x-h)^2+(y-k)^2=r^2\)be a circle whose center is at A(h,k) and let \(P(x_1,y_1)\) be a given point on C.
   Then
   Equation of straight line passing through P is
   \( y-y_1=m(x-x_1)\)
   Now, the slope of the tangent to the circle is
   \(m_t=\frac{dy}{dx}\)
   or \( m_t=-\frac{x_1-h}{y_1-k}\) at \(P(x_1,y_1)\)
   Again, the slope of the normal line to the circle is
   \(m_n=\frac{y_1-k}{x_1-h}\)
   Hence, equation of normal line is
   \( y-y_1=m(x-x_1)\)
   or \( y-y_1=\frac{y_1-k}{x_1-h} (x-x_1)\) Standard form
   This completes the proof

3. Equation of normal line to the circle \(x^2+y^2+2gx+2fy+c=0\)
   

   Let \(C: x^2+y^2+2gx+2fy+c=0\) be a circle whose center and let \(P(x_1,y_1)\) be a given point on C.
   Then
   \(x_1^2+y_1^2+2gx_1+2fy_1+c=0\) (i)
   Also let \(Q(x_2,y_2)\) be a nearby point on C, then
   \( x_2^2+y_2^2+2gx_2+2fy_2+c=0\) (ii)
   Subtracting (i) from (ii), we get
   \( (x_2^2-x_1^2)+(y_2^2-y_1^2)+2g(x_2-x_1)+2f(y_2-y_1)=0\)
   or \( (x_2-x_1)(x_1+x_2+2g)+(y_2-y_1)(y_1+y_2+2f)=0\)
   or \( \frac{y_2-y_1}{x_2-x_1} =- \frac{x_1+x_2+2g}{y_1+y_2+2f} \)
   We know that, tangent is the limiting position of secant, so in the tangent line, the points \(P(x_1,y_1)\) and \(Q(x_2,y_2)\) coincides
   Now, the slope of tangent to the circle at \(P(x_1,y_1)\) is
   \(m=- \frac{x_1+x_2+2g}{y_1+y_2+2f} \)
   or \( m=-\frac{2x_1+2g}{2y_1+2f}\)
   or \( m=-\frac{x_1+g}{y_1+f} \)
   Again, the slope of the normal line to the circle is
   \(m_n=\frac{y_1+f}{x_1+g}\)
   Hence, the equation of normal line to the circle at \(P(x_1,y_1)\) is
   \( y-y_1=m(x-x_1)\)
   or \( y-y_1=\frac{y_1+f}{x_1+g} (x-x_1)\) Standard form
   This completes the proof

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Length of the tangent from external point

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Let \(C: x^2+y^2+2gx+2fy+c=0\) be a circle and \(P(x_1,y_1)\) ba an external point from which a tangent PT is drawn to the circle C.
Then
center of the circle A(h,k)= (-g,-f)
radius of the circle r= \(\sqrt{g^2+f^2-c} \)

In this case
\(PT^2=PA^2-AT^2 \)
or\(PT^2=(x_1+g)^2 +(y_1+f)^2- (g^2+f^2-c)\)
or\(PT^2=x_1^2+y_1^2+2gx_1+2fy_1+c\)
Hence, the length of the tangent from external point is
\(PT=\sqrt{x_1^2+y_1^2+2gx_1+2fy_1+c} \)
NOTE:
The length of the tangent from external point \(P(x_1,y_1)\) to the circle \(C: x^2+y^2=r^2\) is
\(PT=\sqrt{x_1^2+y_1^2-r^2} \)

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Exercise 1

1. Find the equation of tangent and normal to given circle at given point
   1. x² + y² = 25 at (3, -4)
      

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      Solution
      Given equation of circle is
      \(x^2+y^2=25\)
      The radius of the circle is
      \(r^2=25\)
      Given point is
      \((x_1,y_1)=(3,4) \)
      Now, the equation of tangent is
      \( y-y_1 = - \frac{x_1}{y_1} (x-x_1)\)
      \( y+4 = - \frac{3}{(-4)} (x-3) \)
      \(3x-4y=25\)
      

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      Next, the equation of normal is (perpendicular to the tangent and passes through origin)
      \(4x+3y=0\)
      
   2. \( x^2+y^2=4 \) at \( (1,\sqrt{3} ) \)
      

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      Solution
      Given equation of circle is
      \(x^2+y^2=4\)
      The radius of the circle is
      \(r^2=4\)
      Given point is
      \((x_1,y_1)= (1,\sqrt{3} ) \)
      Now, the equation of tangent is
      \( y-y_1 = - \frac{x_1}{y_1} (x-x_1)\)
      \( y-\sqrt{3} = - \frac{1}{\sqrt{3}} (x-1)\)
      \(x+\sqrt{3}y=4\)
      

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      Next, the equation of normal is (perpendicular to the tangent and passes through origin)
      \(\sqrt{3}x-y=0\)
      
   3. \(x^² + y^² =4 \) at \( (2 \cos \theta, 2 \sin \theta) \)
      

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      Solution
      Given equation of circle is
      \(x^2 + y^2 = 4\)
      The radius of the circle is
      \(r^2 = 4 \)
      Given point is
      \((x_1, y_1) = (2 \cos \theta, 2 \sin \theta) \)
      Now, the equation of tangent at \((x_1, y_1)\) is
      \( y-y_1 = - \frac{x_1}{y_1} (x-x_1)\)
      Substitute \(x_1\) and \(y_1\) into the equation
      \( y-2 \sin \theta= - \frac{2 \cos \theta}{2 \sin \theta} (x-2 \cos \theta)\)
      \(x \cos \theta + y \sin \theta = 2 \)
      

      ---

      Next, the equation of normal is (perpendicular to the tangent and passes through origin)
      \(x \sin \theta-y \cos \theta=0\)
      
   4. \( x^2+y^2=8 \) at \( (2,2) \)
      

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      Solution
      Given equation of circle is
      \(x^2 + y^2 = 8\)
      The radius of the circle is
      \(r^2 = 8 \)
      Given point is
      \((x_1, y_1) = (2, 2) \)
      Now, the equation of tangent at \((x_1, y_1)\) is
      \( y-y_1 = - \frac{x_1}{y_1} (x-x_1)\)
      Substitute \(x_1\) and \(y_1\) into the equation
      \( y-2 = - \frac{2}{2} (x-2)\)
      Simplify
      \( x + y = 4 \)
      

      ---

      Next, the equation of normal is (perpendicular to the tangent and passes through origin)
      \(y-x=0\)
      
   5. \( x^2+y^2=36 \) at \( (-6,0) \)
      

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      The radius of the circle is
      \(r^2 = 36 \implies r = 6\)
      Given point is
      \((x_1, y_1) = (-6, 0) \)
      Now, the equation of tangent at \((x_1, y_1)\) is
      \( y-y_1 = - \frac{x_1}{y_1} (x-x_1)\)
      Substitute \(x_1\) and \(y_1\) into the equation
      \( y-0 = - \frac{-6}{0} (x-6)\)
      Simplifying gives
      \(x-6=0 \)
      

      ---

      Next, the equation of normal is (perpendicular to the tangent and passes through origin)
      \(y=0\)
      
   6. \(x^² + y^² + 2 x + 4 y —20 = 0 \) at (3,1)
      

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      Solution
      Given equation of circle is
      \(x^² + y^² + 2 x + 4 y —20 = 0 \)
      The center of the circle is
      \((h,k)=(-g,-f)=(-1,-2)\)
      The radius of the circle is
      \(r^2=g^2+f^2-c=25\)
      Given point is
      \((x_1,y_1)= (3,1) \)
      Now, the equation of tangent is
      \( y-y_1 = - \frac{x_1-h}{y_1-k} (x-x_1)\)
      \( y-1 = - \frac{3+1}{1+2} (x-3)\)
      \(4x+3y=15\)
      

      ---

      Next, the equation of normal is (perpendicular to the tangent)
      \(3x-4y=k\)
      The normal line passes through center \((h,k)=(-g,-f)=(-1,-2)\), so the equation is
      \(3x-4y=5\)
      
   7. \(x^² + y^² -6x-8y-4 =0\) at (8,6)
      

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      Solution
      Given equation of circle is
      \(x^2 + y^2 - 6x - 8y - 4 = 0\)
      To find the center and radius, we rewrite the equation in standard form
      \((x - 3)^2 + (y - 4)^2 = 29\)
      The center of the circle is
      \((h, k) = (3, 4)\)
      The radius of the circle is
      \(r^2 = 29 \)
      Given point is
      \((x_1, y_1) = (8, 6) \)
      Now, the equation of tangent is
      \( y - y_1 = -\frac{x_1 - h}{y_1 - k}(x - x_1) \)
      Substitute \(h = 3\), \(k = 4\), \(x_1 = 8\), and \(y_1 = 6\) into the equation
      \( y - 6 = -\frac{8 - 3}{6 - 4}(x - 8) \)
      \( y - 6 = -\frac{5}{2}(x - 8) \)
      \( 5x + 2y = 52 \)
      

      ---

      Next, the equation of normal is (perpendicular to the tangent)
      \(2x-5y=k\)
      The normal line passes through center \((h,k)=(3, 4)\), so the equation is
      \(2x-5y+14=0\)
      
   8. \(x^² + y^² -3x+10y-15=0 \) at (4,-11)
      

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      Solution
      Given equation of circle is
      \(x^2 + y^2 - 3x + 10y - 15 = 0\)
      To find the center and radius, we rewrite the equation in standard form
      \((x - \frac{3}{2})^2 + (y + 5)^2 = \frac{139}{4} \)
      The center of the circle is
      \((h, k) = (\frac{3}{2}, -5)\)
      The radius of the circle is
      \(r^2 = \frac{139}{4} \)
      Given point is
      \((x_1, y_1) = (4, -11) \)
      Now, the equation of tangent is
      \( y - y_1 = -\frac{x_1 - h}{y_1 - k}(x - x_1) \)
      Substitute \(h = \frac{3}{2}\), \(k = -5\), \(x_1 = 4\), and \(y_1 = -11\) into the equation
      \( y + 11 = -\frac{4 - \frac{3}{2}}{-11 + 5}(x - 4) \)
      \( y + 11 = -\frac{\frac{5}{2}}{-6}(x - 4) \)
      \( y + 11 = \frac{5}{12}(x - 4) \)
      \( 5x - 12y = 152 \)
      

      ---

      Next, the equation of normal is (perpendicular to the tangent)
      \(12x+5y=k\)
      The normal line passes through center \((h,k)=(\frac{3}{2}, -5)\), so the equation is
      \(12x+5y+7=0\)
      
   9. \(x^2+y^2-8x-2y+12=0\) at (x, -1)

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      Solution
      Given equation of circle is
      \(x^2 + y^2 - 3x + 10y - 15 = 0\)
      To find the center and radius, we rewrite the equation in standard form, then
      The center of the circle is
      \((h, k) = (4,1)\)
      The radius of the circle is
      \(r^2 = 5 \)
      Given point is
      \((x_1, y_1) = (x,-1) \)
      Now, the equation of tangent is
      \( y - y_1 = -\frac{x_1 - h}{y_1 - k}(x - x_1) \)
      \( y +1 = -\frac{x - 4}{-1 -1}(x - x) \)
      \( y + 1 = 0 \)
      

      ---

      Next, the equation of normal is (perpendicular to the tangent) is
      \(x+k=0\)
      The normal line passes through center \((h,k)=(4,1) \), so the equation is
      \(x-4=0\)
      
2. Find the point where the tangent to the circle \(x^2+y^2=225\) at (9, 12) crosses the x-axis. Ans: (25,0)
   

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   Solution
   Given equation of circle is
   \(x^2+y^2=225\)
   The radius of the circle is
   \(r^2=225\)
   Given point is
   \((x_1,y_1)=(9,12) \)
   Now, the equation of tangent is
   \( y-y_1 = - \frac{x_1}{y_1} (x-x_1)\)
   \( y-12 = - \frac{9}{12} (x-9)\)
   \(9x + 12y = 225\)
   We know that any line cross the x-axis at (x,0), so the point where the tangent cross the x-axis is
   \(9x + 12y = 225\)
   \(9x+12(0)=225\)
   \(x=25\)
   Therefore, the point is
   (25,0)
3. Find the point where the tangent to the circle \(x^2+y^2=25\) at (2, 4) crosses the x-axis. Ans: (10,0)
   

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   Solution
   Given equation of circle is
   \(x^2+y^2=20\)
   The radius of the circle is
   \(r^2=20\)
   Given point is
   \((x_1,y_1)=(2,4) \)
   Now, the equation of tangent is
   \( y-y_1 = - \frac{x_1}{y_1} (x-x_1)\)
   \( y-4 = - \frac{2}{4} (x-2)\)
   \(x+2y=10\)
   We know that any line cross the x-axis at (x,0), so the point where the tangent cross the x-axis is
   \(x+2y=10\)
   \(x+2(0)=10\)
   \(x=10\)
   Therefore, the point is
   (10,0)
4. Find the following
   1. Find the equation of tangent and normal to \(x^² + y^² =40 \) at the points whose (i) absciassa is 2 (ii) ordinate is -6

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   Solution
   
   1. Abscissa is 2
      Given the abscissa (x-coordinate) is 2, we find the corresponding ordinate (y-coordinate) using the circle equation, so
      \(x^2 + y^2 = 40\)
      or \(2^2 + y^2 = 40\)
      or \(4 + y^2 = 40\)
      or \(y^2 = 36\)
      or \(y = \pm 6\)
      Now, equation of the tangent to the circle at \((x_1, y_1)=(2, 6)\) is
      \(xx_1 + yy_1 = 40\)
      \(2x + 6y = 40\)
      \(x + 3y = 20\)
      Again, equation of the normal to the circle at \((x_1, y_1)=(2, 6)\) is , and which pass through (0,0) is
      \(3x-y=0\)
      

      ---

      Next, equation of the tangent to the circle at \((x_1, y_1)=(2,- 6)\) is
      \(xx_1 + yy_1 = 40\)
      \(2x -6y = 40\)
      \(x -3y = 20\)
      Also, equation of the normal to the circle at \((x_1, y_1)=(2, -6)\) is , and which pass through (0,0) is
      \(3x+y=0\)
      

      ---

      ---

   2. Ordinate is -6
      Given the ordinate (y-coordinate) is -6, we find the corresponding abscissa (x-coordinate) using the circle , so
      \(x^2 + y^2 = 40\)
      or \(x^2 + (-6)^2 = 40\)
      or \(x^2 +36= 40\)
      or \(x^2 = 4\)
      or \(x = \pm 2\)
      So the points are (2, -6) and (-2, -6)
      Now, equation of the tangent to the circle at \((x_1, y_1)=(-2,- 6)\) is
      \(xx_1 + yy_1 = 40\)
      \(-2x -6y = 40\)
      \(x +3y = 20\)
      Also, equation of the normal to the circle at \((x_1, y_1)=(-2, -6)\) is , and which pass through (0,0) is
      \(3x-y=0\)
      
   2. Find the equation of tangent to the circle \(2x^² + 2y^² =9 \) which makes angle 45 degree with x-axis
      

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      Solution
      Given equation of circle is
      \(2x^2 + 2y^2 = 9\)
      \(x^2 + y^2 = \frac{9}{2}\)
      Now, the center is \(C=(0, 0)\) and radius \(r^2=\frac{9}{2}\)
      Given that, the tangent to the circle that makes an angle of 45 degrees with the x-axis , so
      slope \(m\) of tangent is
      \( m=\tan(45) = 1 \)
      Hence, equation of a tangent line is
      \(y = x + c\)
      To find \(c\), we use the condition of tangency fro the line \(y=mx+c\) to the circle \(x^2+y^2=r^2\) with
      \(c=0\)
      \(r^2=\frac{9}{2}\)
      \( m=1\)
      The condition is
      \(c^2 = \pm r^2 (1 +m^2)\)
      \(c^2 = \pm \frac{9}{2} (1 +1)\)
      \(c = \pm 3\)
      Therefore, the equations of the tangents are
      \(y = x \pm 3\)
      
   3. Find the equation of normal to the circle \(2x^² + 2y^² =9 \) which makes angle 45 degree with x-axis

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      Solution
      Given equation of circle is
      \(2x^2 + 2y^2 = 9\)
      \(x^2 + y^2 = \frac{9}{2}\)
      Now, the center is \(C=(0, 0)\) and radius \(r^2=\frac{9}{2}\)
      Given that, the normal to the circle that makes an angle of 45 degrees with the x-axis , so
      slope \(m\) of normal is
      \( m=\tan(45) = 1 \)
      Hence, equation of a normal line (which passes through origin (0,0) is
      \(y = mx\)
      \(y = x \)
      
5. Find the following equation of tangent to the circle
   1. \(x^2+y^2=4\) which are parallel to \(3x+4y-5=0\)

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      Solution
      Given Circle is
      \(x^2 + y^2 = 4\)
      In which
      center \(C = (0, 0)\) and radius \(r = 2\)
      Now, the equation of a tangent line parallel to \(3x + 4y - 5 = 0\) is
      \(3x + 4y + k = 0\)
      We will find the value of \(k\) using the condition of tangency, which is
      \(d = \pm \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}\)
      \(2 = \pm \frac{|3(0) + 4(0) + k|}{\sqrt{3^2 + 4^2}}\)
      \(2 = \pm \frac{|k|}{5}\)
      \(10 = \pm |k|\)
      \(k = 10\) or \(k = -10\)
      Hence, the equations of the tangents are
      \(3x + 4y + 10 = 0\)
      \(3x + 4y - 10 = 0\)
      
   2. \(x^2+y^2=5\) which are parallel to \(x+2y=0\)

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      Given Circle is
      \(x^2 + y^2 = 5\)
      In which
      center \(C = (0, 0)\) and radius \(r = \sqrt{5}\)
      Now, the equation of a tangent line parallel to \(x + 2y = 0\) is
      \(x + 2y + k = 0\)
      To find the value of \(k\), we use the condition of tangency, which is
      \(d = \pm \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}\)
      \(\sqrt{5} = \pm \frac{|1(0) + 2(0) + k|}{\sqrt{1^2 + 2^2}}\)
      \(\sqrt{5} = \pm \frac{|k|}{\sqrt{5}}\)
      5 = \pm |k|\)
      \(k = \sqrt{5}\) or \(k = -\sqrt{5}\)
      Hence, the equations of the tangents are
      \(x + 2y + \sqrt{5} = 0\)
      \(x + 2y - \sqrt{5} = 0\)
      
   3. \(x^2+y^2-6x+4y=12\) which are parallel to \(4x+3y+5=0\)
      

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      Solution
      Given Circle is
      \(x^2 + y^2 - 6x + 4y = 12\)
      In which
      center \(C=(3, -2)\) and radius \(r=5\)
      Now, equation of a tangent line parallel to \(4x + 3y + 5 = 0\) is
      \(4x + 3y + k = 0\)
      To find the value of \(k\), we use condition of tangentcy, which is
      \(d = \pm \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}\)
      \(5 = \pm \frac{|4(3) + 3(-2) + k|}{\sqrt{4^2 + 3^2}}\)
      \(5 = \pm \frac{|6+k|}{5}\)
      \(25 = \pm (6+k)\)
      k=19 or -31
      Hence, the equation of tangents are
      \(4x + 3y + 19 = 0\)
      \(4x + 3y - 31 = 0\)
      
   4. \(x^2+y^2-2x-4y-4=0\) which are parallel to \(3x-4y=1\)

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      Solution
      Given Circle is
      \(x^2 + y^2 - 2x - 4y - 4 = 0\)
      In which
      center \(C = (1, 2)\) and radius \(r = 3\)
      Now, the equation of a tangent line parallel to \(3x - 4y = 1\) is
      \(3x - 4y + k = 0\)
      To find the value of \(k\), we use the condition of tangency, which is
      \(d = \pm \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}\)
      \(3 = \pm \frac{|3(1) - 4(2) + k|}{\sqrt{3^2 + 4^2}}\)
      \(3 = \pm \frac{|3 - 8 + k|}{5}\)
      \(3 = \pm \frac{|k - 5|}{5}\)
      \(15 = \pm |k - 5|\)
      \(k - 5 = 15\) or \(k - 5 = -15\)
      \(k = 20\) or \(k = -10\)
      Hence, the equations of the tangents are
      \(3x - 4y + 20 = 0\)
      \(3x - 4y - 10 = 0\)
      
6. Show that
   1. tangent to the circle \(x^2+y^2=100\) at the points (6,8) and (8,-6) are perpendicular to each other

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      Solution
      Given Circle is
      \(x^2 + y^2 = 100\)
      In which
      center \(C = (0, 0)\) and radius \(r = 10\)
      Given points of tangency are (6,8) and (8,-6).
      First, find the slope of the tangents at these points.
      The tangent to the circle \(x^2 + y^2 = a^2\) at the point \((x_1, y_1)\) has slope:
      \(m=-\frac{x_1}{y_1}\)
      For the point (6,8):
      The slope is \(m_1=-\frac{6}{8}\)
      For the point (8,-6):
      The slope is \(m_2=-\frac{8}{-6}=\frac{8}{6}\)
      Check the condition:
      \(m_1 \times m_2=-1\)TRUE
      Therefore, the tangents are perpendicular to each other.
      
   2. tangent to the circle \(x^2+y^2+4x+8y+2=0\) at the points (1,-1) and (-5,-7) are parallel to each other

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      Solution
      Given Circle is
      \(x^2+y^2+4x+8y+2=0\)
      In which
      center \(C = (h,k)=(-2, -4)\) and radius \(r = \sqrt{f^2+g^2-c}=\sqrt{18}\)
      Given points of tangency are (1,-1) and (-5,-7)
      First, find the slope of the tangents at these points.
      The tangent to the circle at the point \((x_1, y_1)\) has slope:
      \(m=-\frac{x_1-h}{y_1-k}\)
      For the point (1,-1):
      The slope is \(m_1= -\frac{1+2}{-1+4}=-1\)
      For the point (-5,-7):
      The slope is \(m_2=-\frac{-5+2}{-7+4}=-1\)
      Check the condition:
      \(m_1 = m_2\)TRUE
      Therefore, the tangents are parallel to each other.
      
7. Find the equation of the circle whose center is (h,k) and which passes through the origin and prove that the equation of the tangent at the origin is \(hx+ky=0\)

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   Solution
   The circle whose center is \((h, k)\) and which passes through the origin is
   \((x - h)^2 + (y - k)^2 = r^2\)
   Since the circle passes through the origin \((0, 0)\), substitute \((0, 0)\) into the equation, we get
   \((0 - h)^2 + (0 - k)^2 = r^2\)
   \(h^2 + k^2 = r^2\)
   Thus, the equation of the circle is:
   \((x - h)^2 + (y - k)^2 = h^2 + k^2\)
   \(x^2 + y^2 - 2hx - 2ky = 0\)
   Next
   The equation of the tangent to the circle is
   \(xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0\)
   At the origin \((x_1,y_1)=(0, 0)\), the equation of the tangent is, keeping g=-2, h=-2
   \(hx + ky = 0\)
   

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Exercise 2

1. Show that the line \(3x-4y=25\) and the circle \(x^2+y^2=25\) intersect at two coincide points

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   Solution
   Given equation of circle is
   \(x^2 + y^2 = 25\)(i)
   Now, the center is \(C=(0, 0)\) and radius \(r^2=25\)
   Given that, tangent line is
   \(3x-4y=25\)
   \(3x-4y-25=0\) (ii)
   Here, the condition that (ii) to be tangent (intersect at two coincide points) to (i) is
   \(c^2 = \pm r^2 (a^2 +b^2)\)
   \((-25)^2 = \pm 25 (3^2 +4^2) \) TRUE
   It shows that line (ii) is tangent to circle (i).
2. Prove that the line \(5x+12y+78=0\) is tangent to the circle \(x^2+y^2=36\)

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   Solution
   Given equation of circle is
   \(x^2 + y^2 = 36\)(i)
   Now, the center is \(C=(0, 0)\) and radius \(r^2=25\)
   Given that, tangent line is
   \(5x+12y+78=0\) (ii)
   Here, the condition that (ii) to be tangent (intersect at two coincide points) to (i) is
   \(c^2 = \pm r^2 (a^2 +b^2)\)
   \((78)^2 = \pm 36 (5^2 +12^2) \)
   \((78)^2 = \pm 6^2 (13^2) \) TRUE
   It shows that line (ii) is tangent to circle (i).
3. Prove that the tangent to the circle \(x^2+y^2=5\) at the point (1,-2) also touches the circle \(x^2+y^2-8x+6y+20=0\) and find the point of contact.

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   Solution
   Given equation of circle is
   \(x^2+y^2=5\) (i)
   The radius of the circle is
   \(r^2=5\)
   Given point is
   \((x_1,y_1)=(1,-2) \)
   Now, the equation of tangent to (i) is
   \( xx_1+yy_1=r^2\)
   \(x-2y=5\) (ii)
   

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   Next, given circle is
   \(x^2+y^2-8x+6y+20=0\) (iii)
   The center of the circle (iii) is
   \(A(h, k) = (4,-3)\)
   The radius of the circle (iii) is
   \(r^2 =4^2+3^2-20=5 \)
   We have to show that line (ii) is also to the circle (iii)
   Here, the condition that (ii) to be tangent to (iii) is
   \(d = \pm \frac{Ax+By+C}{\sqrt{A^2+B^2}}\)
   \(\sqrt{5} = \pm \frac{1(h)-2(k)-5}{\sqrt{1^2+2^2}}\)
   \(\sqrt{5} = \pm \frac{1(4)-2(-3)-5}{\sqrt{5}}\)
   \(\sqrt{5} = \pm \sqrt{5} \)TRUE
   So, the tangent to the circle \(x^2+y^2=5\) at the point (1,-2) also touches the circle \(x^2+y^2-8x+6y+20=0\)
   

   ---

   The point of contact to (iii) is obtained by the intersection of tangent and normal line to circle (iii), i.e.,
   The equation of tangent to (iii) is
   \(x-2y=5\) (iv)
   The equation of normal to (iii) is
   \(2x+y=k\)
   It passes through (4,-3), so k=5, hence, the equation of normal line is
   \(2x+y=5\) (v)
   Solving (iv) and (v), we get
   point of contact=(3,-1)
   
4. Prove that the line \(y=x+a\sqrt{2}\) touches the circle \(x^2+y^2=a^2\) and find the point of contact.

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   Solution
   Given line is
   \(y=x+a\sqrt{2}\) (i)
   

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   Next, given circle is
   \(x^2+y^2=a^2\) (ii)
   We have to show that line (i) is tangent to the circle (ii)
   Here, the condition that (i) to be tangent to (ii) is
   \(c^2 = \pm r^2 (1+m^2)\)
   \(2a^2 = \pm a^2 (1+1)\)TRUE
   This completes the first part
   

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   The point of contact to (i) and (ii) is obtained by the intersection of tangent and normal line to circle (ii), i.e.,
   The equation of tangent to (ii) is
   \(y=x+a\sqrt{2}\) (iii)
   The equation of normal to (ii) is (which passes through origin (0,0))
   \(x+y=0\) (iv)
   Solving (iii) and (iv), we get
   point of contact=\( \left ( \frac{a}{\sqrt{2}}, -\frac{a}{\sqrt{2}}\right ) \)
   
5. Find the value of k so that
   1. the line \(4x+3y+k=0\) may touches the circle \(x^2+y^2-4x+10y+4=0\)

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      Solution
      Given Circle is
      \(x^2+y^2-4x+10y+4=0\)
      In which
      center \(C =(-f,-g)= (2, 5)\) and radius \(r = \sqrt{f^2+g^2-c}=5\)
      Given line is
      \(4x+3y+k=0\)
      In which
      \(A=4,B=3,C=k\)
      We will find the value of \(k\) using the condition of tangency, which is
      \(d = \pm \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}\)
      \(5 = \pm \frac{|4(2) + 3(5) + k|}{\sqrt{4^2 + 3^2}}\)
      \(5 = \pm \frac{|23+k|}{5}\)
      \(25 = \pm |23+k|\)
      \(k = 2\) or \(k = -48\)
      
   2. the line \(2x-y+4k=0\) touches the circle \(x^2+y^2-2x-2y-3=0\)

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      Solution
      Given Circle is
      \(x^2+y^2-2x-2y-3=0\)
      In which
      center \(C =(-f,-g)= (1, 1)\) and radius \(r = \sqrt{f^2+g^2-c}=\sqrt{5}\)
      Given line is
      \(2x-y+4k=0\)
      In which
      \(A=2,B=-1,C=4k\)
      We will find the value of \(k\) using the condition of tangency, which is
      \(d = \pm \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}\)
      \(\sqrt{5} = \pm \frac{|2(1) -1(1) + 4k|}{\sqrt{2^2 + 1^2}}\)
      \(5 = \pm \frac{|4k+1|}{\sqrt{5}}\)
      \(5 = \pm |4k+1|\)
      \(k = 1\) or \(k = -3/2\)
      
6. Find the condition that
   1. the line \(px+qy=r\) is tangent to the circle \(x^2+y^2=a^2\)

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      Solution
      Given Circle is
      \(x^2+y^2=a^2\)
      In which
      center \(C =(h,k)= (0,0)\) and radius \(r = a\)
      Given line is
      \(px+qy=r\)
      In which
      \(A=p,B=q,C=-r\)
      The condition of given line to be tangency to the circle is
      \(d = \pm \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}\)
      \(a = \pm \frac{|p(0) +q(0) -r|}{\sqrt{p^2 + q^2}}\)
      \(a = \pm \frac{-r}{\sqrt{p^2 + q^2}}\)
      \(a^2 = \pm \frac{r^2}{\sqrt{p^2 + q^2}}\)
      
   2. the line \(lx+my+n=0\) is tangent to the circle \(x^2+y^2+2gx+2fy+c=0\)

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      Solution
      Given Circle is
      \(x^2 + y^2 + 2gx + 2fy + c = 0\)
      In which
      center \(C = (-g, -f)\) and radius \(r = \sqrt{g^2 + f^2 - c}\)
      Given line is
      \(lx + my + n = 0\)
      In which
      \(A = l, B = m, C = n\)
      The condition for the given line to be tangent to the circle is
      \(d = \pm \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}\)
      \(r = \pm \frac{|l(-g) + m(-f) + n|}{\sqrt{l^2 + m^2}}\)
      \(r = \pm \frac{| -lg - mf + n|}{\sqrt{l^2 + m^2}}\)
      \(\sqrt{g^2 + f^2 - c} = \pm \frac{| -lg - mf + n|}{\sqrt{l^2 + m^2}}\)
      \((g^2 + f^2 - c) = \pm \frac{(-lg - mf + n)^2}{(l^2 + m^2)}\)
      
   3. the line \(lx+my+n=0\) is normal to the circle \(x^2+y^2+2gx+2fy+c=0\)

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      Solution
      Given Circle is
      \(x^2 + y^2 + 2gx + 2fy + c = 0\)
      In which
      center \(C = (-g, -f)\) and radius \(r = \sqrt{g^2 + f^2 - c}\)
      Given line is
      \(lx + my + n = 0\)
      In which
      \(A = l, B = m, C = n\)
      For the line to be normal to the circle, it must pass through the center of the circle
      Substitute \(x = -g\) and \(y = -f\) into the line equation:
      \(l(-g) + m(-f) + n = 0\)
      \(-lg - mf + n = 0\)
      \(n = lg + mf\)
      Thus, the line \(lx + my + n = 0\) is normal to the circle \(x^2 + y^2 + 2gx + 2fy + c = 0\) if \(n = lg + mf\).
      
   4. the circle \(x^2+y^2+2gx+2fy+c=0\) touches the (i) x-axis (ii) y-axis

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      Solution
      Given Circle is
      \(x^2 + y^2 + 2gx + 2fy + c = 0\)
      In which
      center \(C = (-g, -f)\) and radius \(r = \sqrt{g^2 + f^2 - c}\)
      (i) To determine the condition for the circle to touch the x-axis:
      The circle will touch the x-axis if the distance from the center to the x-axis is equal to the radius.
      Distance from center to x-axis = \(|f|\)
      Radius \(r = \sqrt{g^2 + f^2 - c}\)
      Thus, \(|f| = \sqrt{g^2 + f^2 - c}\)
      Square both sides:
      \(f^2 = g^2 + f^2 - c\)
      \(c = g^2\)
      (ii) To determine the condition for the circle to touch the y-axis:
      The circle will touch the y-axis if the distance from the center to the y-axis is equal to the radius.
      Distance from center to y-axis = \(|g|\)
      Radius \(r = \sqrt{g^2 + f^2 - c}\)
      Thus, \(|g| = \sqrt{g^2 + f^2 - c}\)
      Square both sides:
      \(g^2 = g^2 + f^2 - c\)
      \(c = f^2\)
      
7. If the line \(lx+my=1\) touches the circle \(x^2+y^2=a^2\), prove that the point \((l,m)\) lies on a circle whose radius is \(\frac{1}{a}\)

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   Solution
   Given Circle is
   \(x^2 + y^2 = a^2\)
   In which
   center \(C = (0, 0)\) and radius \(r = a\)
   Given line is
   \(lx + my = 1\)
   For the line to touch the circle, the condition is
   \(d = \frac{|l(0) + m(0) - 1|}{\sqrt{l^2 + m^2}}\)
   \(a = \frac{| -1|}{\sqrt{l^2 + m^2}}\)
   \(a^2 = \frac{1}{l^2 + m^2}\)
   \(l^2 + m^2 = \frac{1}{a^2}\)
   Thus, the point \((l, m)\) lies on a circle \(x^2 + y^2 = \frac{1}{a^2}\), whose radius \(\frac{1}{a}\)
8. Do the following
   1. Find the condition for the two circles \(x^2+y^2=a^2\) and \( (x-c)^2+y^2=b^2\) to touch (i) externally and (ii) internally
      

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      Solution
      The two circles are
      Circle 1: \(x^2 + y^2 = a^2\)
      Circle 2: \((x - c)^2 + y^2 = b^2\)
      So, the centers and radii of the circles are
      Circle 1: Center \(c_1: (0, 0)\), Radius \(r_1= a\)
      Circle 2: Center \(c_2: (c, 0)\), Radius \(r_2=b\)
      

      • The two circles touch externally if the distance between their centers is equal to the sum of their radii \(d(c_1,c_2)=r_1+r_2\)
        \(c = a + b\)
        
      • The two circles touch internally if the distance between their centers is equal the absolute difference of their radii
        \(d(c_1,c_2)=|r_1-r_2|\)
        \(c = a - b\)
        
   2. Prove that the two circles \(x^2+y^2+2ax+c^2=0\) and \(x^2+y^2+2by+c^2=0\) touch if \(\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}\)
      

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      Solution
      Given the two circles are
      Circle 1: \(x^2 + y^2 + 2ax + c^2 = 0\)
      Circle 2: \(x^2 + y^2 + 2by + c^2 = 0\)
      Here
      The center of Circle 1 is
      \(C_1:(-a, 0)\) and its radius is \(r_1=\sqrt{a^2 - c^2}\).
      The center of Circle 2 is
      \(C_2:(0, -b)\) and its radius is \(r_2=\sqrt{b^2 - c^2}\).
      Next
      The distance \(d\) between the centers of the circles is
      \(d=d(C_1,C_2)\)
      \(d = \sqrt{(-a - 0)^2 + (0 - (-b))^2} \)
      \(d = \sqrt{a^2 + b^2}\)
      Now, two circles touch, if the distance between their centers is equal to the sum of their radii, so
      \(d(C_1,C_2) =r_1+r_2\)
      \(\sqrt{a^2 + b^2} = \sqrt{a^2 - c^2} + \sqrt{b^2 - c^2}\)
      Square both sides, we get
      \(a^2 + b^2 = (\sqrt{a^2 - c^2} + \sqrt{b^2 - c^2})^2\)
      \(a^2 + b^2 = (a^2 - c^2) + 2\sqrt{(a^2 - c^2)(b^2 - c^2)} + (b^2 - c^2)\)
      \(a^2 + b^2 = a^2 + b^2 - 2c^2 + 2\sqrt{(a^2 - c^2)(b^2 - c^2)}\)
      \(c^2 = \sqrt{(a^2 - c^2)(b^2 - c^2)}\)
      Square both sides again, we get
      \(c^4 = (a^2 - c^2)(b^2 - c^2)\)
      \(c^4 = a^2b^2 - a^2c^2 - b^2c^2 + c^4\)
      \(a^2b^2 = a^2c^2 + b^2c^2\)
      Divide both sides by \(a^2b^2c^2\), we get
      \(\frac{1}{c^2} = \frac{1}{a^2} + \frac{1}{b^2}\)
      This completes the solution

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Exercise 3

1. Find the equation to the pair of tangents drawn from the origin to the circle \(x^2+y^2-4x-4y+7=0\)
   

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   Solution
   Given external point is
   \((x_1,y_1)=(0,0)\)
   Given equation of circle is
   \(S:x^2+y^2-4x-4y+7=0\)
   We know
   For the circle is
   \(S:x^2+y^2+2gx+2fy+c=0\)
   For the length of tangent to the circle from \((x_1,y_1)\) is
   \(S_1:x_1^2+y_1^2+2gx_1+2fy_1+c=0\)
   For equation of chord of contact is
   \(T:xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0\)
   The pair of tangents drawn from the origin to the circle is
   \(T^2(0,0)=S(x,y)S_1(0,0)\)
   or \((2x+2y+7)^2=(x^2+y^2-4x-4y+7)(7)\)
   or \(3x^2-8xy+3y^2=0\)
   
2. Find the equation of tangents drawn from the point (11,3) to the circle \(x^2+y^2=65\). Also find the angles between the two tangents.

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   Solution
   Given equation of circle is
   \(S: x^2 + y^2 = 65\)
   Given external point is
   \((x_1,y_1)=(11,3)\)
   We know
   Equation of tangent from \((x_1,y_1)=(11,3)\) is
   \(y-y_1=m(x-x_1)\)
   \(y-3=m(x-11)\)
   \(mx-y+(3-11m)=0\)
   We will find the value of \(m\) using the condition of tangency, which is
   \(d = \pm \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}\)
   \(65 = \pm \frac{(3-11m)^2|}{1+m^2}\)
   \(28m^2-33m-28=0\)
   \(m_1 = -4/7 \) or \(m_2 = 7/4\)
   So, the equation of tangents are
   \(y-3=m(x-11)\)
   \(y-3=-\frac{4}{7}(x-11)\) and \(y-3=\frac{7}{4}(x-11)\)
   The angle between is 90, because \( m_1 \times m_-1\)
3. Find the equation of tangents drawn from the origin to the circle \(x^2+y^2+10x+10y+40=0\)

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   Solution
   Given external point is
   \((x_1,y_1)=(0,0)\)
   Given equation of circle is
   \(S:x^2+y^2+10x+10y+40=0\)
   We know
   For the circle is
   \(S:x^2+y^2+2gx+2fy+c=0\)
   For the length of tangent to the circle from \((x_1,y_1)\) is
   \(S_1:x_1^2+y_1^2+2gx_1+2fy_1+c=0\)
   For equation of chord of contact is
   \(T:xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0\)
   The pair of tangents drawn from the origin to the circle is
   \(T^2(0,0)=S(x,y)S_1(0,0)\)
   or \((-5x-5y+40)^2=(x^2+y^2+10x+10y+40)(40)\)
   or \(5(-x-y+8)^2=(x^2+y^2+10x+10y+40)(8)\)
   or \(3x^2-10xy+3y^2=0\)
   or \(x-3y=0\) and \(3x-y=0\)
   

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Exercise 4

1. Determine the length of tangents to the circles
   1. \(x^2+y^2=25\) from (3,5)

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      Solution
      Given external point is
      \((x_1,y_1)=(3,5)\)
      Given equation of circle is
      \(S: x^2 + y^2 - 25 = 0\)
      For the length of tangent to the circle from \((x_1,y_1)\) is
      \(L = \sqrt{x_1^2 + y_1^2 -r^2}\)
      \(L = \sqrt{9 + 25 - 25} \)
      \(L = 3 \)
      
   2. \(x^2+y^2+4x+6y-19=0\) from (6,4)

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      Solution
      Given external point is
      \((x_1,y_1)=(6,4)\)
      Given equation of circle is
      \(S: x^2 + y^2 + 4x + 6y - 19 = 0\)
      For the length of tangent to the circle from \((x_1,y_1)\) is
      \(L = \sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c}\)
      \(L = \sqrt{6^2 + 4^2 + 2 \cdot 2 \cdot 6 + 2 \cdot 3 \cdot 4 - 19} \)
      \(L = \sqrt{36 + 16 + 24 + 24 - 19} \)
      \(L = \sqrt{81} \)
      \(L = 9 \)
      
2. Determine the value of k so that the length of the tangent from (5,4) to the circle \(x^2+y^2+2ky=0\) is 5.

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   Solution
   Given external point is
   \((x_1,y_1)=(5,4)\)
   Given equation of circle is
   \(S: x^2 + y^2 + 2ky = 0\)
   For the length of tangent to the circle from \((x_1,y_1)\) is
   \(L = \sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c}\)
   Substituting \(x_1 = 5\), \(y_1 = 4\), \(g = 0\), \(f = k\), and \(c = 0\):
   \(L = \sqrt{5^2 + 4^2 + 2 \cdot 0 \cdot 5 + 2 \cdot k \cdot 4 + 0} \)
   \(L = \sqrt{25 + 16 + 8k} \)
   \(L = \sqrt{41 + 8k} \)
   Given that the length of the tangent is 5,
   \(5 = \sqrt{41 + 8k} \)
   Squaring both sides,
   \(25 = 41 + 8k \)
   Solving for \(k\),
   \(25 - 41 = 8k \)
   \(-16 = 8k \)
   \(k = -2 \)
   
3. Show that the length of the tangent drawn from any point on the circle \(x^2+y^2+2gx+2fy+c=0\) to the circle \(x^2+y^2+2gx+2fy+c_1=0\) is \(\sqrt{c_1-c}\)

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   Solution
   Consider two circles:
   Circle 1:
   \(S: x^2 + y^2 + 2gx + 2fy + c = 0\)
   Circle 2:
   \(S_1: x^2 + y^2 + 2gx + 2fy + c_1 = 0\)
   Let \((x_1, y_1)\) be any point on Circle 1.
   Then
   The length of the tangent from \((x_1, y_1)\) to Circle 2 is given by:
   \(L = \sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c_1}\)
   Substituting \(x_1^2 + y_1^2 + 2gx_1 + 2fy_1 =-c\):
   \(L = \sqrt{-c + c_1}\)