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
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
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
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
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
Using case 1-case 4, for any number \( n \in Z\) , we have \( [r(\cos\theta + i\sin\theta)]^n =r^n[\cos(n\theta)+i \sin(n\theta)]\)
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
\( W=\sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
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
\( w_1[i=0] \) taking k=0 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
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
\( w_1 [k=0]\) taking k=0 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
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
\( w_1 \) taking k=0 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
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
\( w_1 [k=0]\) taking k=0 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
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
\( w_1 [k=0]\) taking k=0 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
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
\( w_1 [k=0]\) taking k=0 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
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
\( w_1\) taking k=0 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
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
\( w_1\) taking k=0 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
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
\( w_1\) taking k=0 in the formula \( \sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ) \)
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
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
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
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
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
Find the 10th roots of 1 The figure shows 10th root of 1.
Drag the value of k=0,1,2,...,9
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.
Calculate the magnitude ∣Z∣ and angle θ of the original impedance.
Determine the cube root of the original impedance in polar form.
Design a new component with an impedance Zn such that (Zn)^3 matches the original impedance.
Express the new impedance in rectangular form and calculate its magnitude and angle.
If \( \bar{z}\) be the conjugate of a complex number \(z\), prove that \(Arg(\bar{z})=2 \pi- Arg(z)\)
If \(z=\cos \theta +i \sin \theta\), prove that \(z^n-\frac{1}{z^n} = 2 \sin n \theta i\)
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 \( z^{-1} =\frac{1}{r} e^{- i\theta}\)
Proof of the Formula
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.
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.
No comments:
Post a Comment