De Moivre's Theorem

Roots of Complex Numbers

De Moivre's Theorem

For any arbitrary angle $\theta$ and any integer $n$,

$(\cos \theta+ i \sin\theta)^{n}=\cos(n\theta)+i \sin (n\theta)$.

Proof: We will show that theorem is true for all positive integer $n$.

$\begin{aligned} \text{ For } n&=1,\\ \text{ LHS }&=(\cos \theta+i \sin \theta)^{1}\\ &=\cos\theta+i \sin \theta\\ \text{ RHS }&=\cos (1\theta)+i \sin (1\theta)\\ &=\cos \theta+i \sin \theta\\ \text{ LHS }&=\text{ RHS } \end{aligned}$
Suppose that it is true for $n=k$.

$(\cos\theta+i\sin\theta)^{k}=\cos (k\theta)+i \sin (k\theta)$

$\begin{aligned} &\text{ For } n=k+1,\\ &\quad (\cos \theta+i \sin \theta)^{k+1}\\ &=(\cos \theta+i \sin \theta)^{k}(\cos \theta+i \sin \theta)\\ &=(\cos (k\theta)+i \sin (k\theta))(\cos \theta+i \sin \theta)\\ &=\cos (k\theta)\cos \theta-\sin (k\theta)\sin \theta\\ &\quad +i (\sin (k\theta)\cos \theta+\cos (k\theta)\sin \theta)\\ &=\cos (k\theta+\theta)+i \sin (k\theta+\theta)\\ &=\cos (k+1)\theta+i \sin (k+1)\theta \end{aligned}$

It is true for $n=k+1$.

By mathematical induction, the theorem is true for all positive integers $n$.

$\begin{aligned} \text{ For } n&=0,\\ \text{ LHS }&=(\cos \theta+i \sin \theta)^{0}=1\\ \text{ RHS }&=\cos (0\theta)+i \sin (0\theta)\\ &=\cos 0+i \sin 0\\ &=1+i0=1\\ \text{ LHS }&=\text{ RHS } \end{aligned}$
It is true for $n=0$.

Now, we will show that the theorem is true when $n$ is a negative integer.

If $n<0$, write $n=-m$ where $m$ is a positive integer.

$\begin{aligned} \text{ Then }&(\cos\theta+i \sin\theta)^{n}\\ =&(\cos \theta+i \sin \theta)^{-m}\\ =&\frac{1}{(\cos \theta+i \sin \theta)^{m}}\\ =&\frac{1}{\cos m\theta+i \sin m\theta}\\ =&\frac{1}{(\cos m\theta+i \sin m\theta)}\\ &\times\frac{(\cos m\theta-i \ sin m\theta)}{(\cos m\theta-i \ sin m\theta)}\\ =&\frac{(\cos m\theta-i \sin m\theta)}{\cos^{2}m\theta-i^{2} \sin^{2}m\theta}\\ =&\frac{(\cos m\theta-i \sin m\theta)}{\cos^{2}m\theta+ \sin^{2}m\theta}\\ =&\frac{(\cos m\theta-i \sin m\theta)}{1}\\ =&\cos m\theta-i \sin m\theta\\ =&\cos (-n\theta)-i \sin (-n\theta)\\ =&\cos n\theta+i \sin n\theta \end{aligned}$

The theorem is true when $n$ is a negative integer.

Remark: If $n$ is a fraction (i.e, $n=\dfrac{p}{q}, q>1$ and $\dfrac{p}{q}$ is in lowest form), we can show that the theorem is true.

Now, we consider $(\cos \theta+i\sin \theta)^{\frac{1}{q}}$ when $n=\dfrac{1}{q}, q>1$.

$(\cos \theta+i\sin \theta)^{\frac{1}{q}}$ means $q^{th}$ roots of the complex number $\cos \theta+i \sin \theta$.

$\begin{aligned} (\cos \theta&+i\sin \theta)^{\frac{1}{q}}=\cos \phi+i\sin \phi\\ \cos \theta&+i\sin \theta=(\cos \phi+i\sin \phi)^{q}\\ \cos \theta&+i\sin \theta=\cos (q\phi)+i\sin (q\phi)\\ \cos \theta&=\cos (q\phi), \sin \theta=\sin (q\phi)\\ \therefore q\phi&=\theta+2k\pi, k= \text{ integer }\\ \therefore \phi&=\frac{\theta +2k\pi}{q}\\ \end{aligned}$
$\begin{aligned} (\cos \theta+i\sin \theta)^{\frac{1}{q}}=&\cos \left(\dfrac{\theta +2k\pi}{q}\right)\\ &+i\sin \left(\dfrac{\theta +2k\pi}{q}\right) \end{aligned}$
for $k=0, 1, 2,...,(q-1)$.

Let $n$ be a positive integer. The $n^{th}$ roots of the complex number $r(\cos \theta+i\sin \theta)$ are given by

$\sqrt[n]{r}\left(\cos \left(\frac{\theta+2k\pi}{n}\right)+i\sin \left(\frac{\theta+2k\pi}{n}\right)\right)$ for $k=0,1,2,...,(n-1).$

$n^{th}$ roots of unity (unity being the number $1$)

The trigonometric form of $1$ is
$\begin{aligned} 1&=\cos 0+i \sin 0 \quad(\text{ or })\\ 1&=\cos (0+2k\pi) +i \sin (0+2k\pi)\\ \sqrt[n]{1}&=1^{\frac{1}{n}}\\ &=\cos \frac{0+2k\pi}{n}+i\sin \frac{0+2k\pi}{n},\\ k&=0, 1, 2, ... (n-1).\\ &=\cos \frac{2k\pi}{n}+i\sin \frac{2k\pi}{n},\\ k&=0, 1, 2, ... (n-1). \end{aligned}$



Example

(a)  Find the solution to the equation $z^{2}=1$.(That is , find the square roots of $1$.)

Solution

$\begin{aligned} z^{2}= 1&=\cos 0+i\sin 0\\ z=\sqrt[2]{1}&=1^{\frac{1}{2}}\\ &=\cos \frac{0+2k\pi}{2}+i\sin \frac{0+2k\pi}{2},\\ \quad k=0&, 1.\\ k=0& ⇒ z=\cos 0+i\sin 0=1\quad \text{ and }\\ k=1& ⇒ z=\cos \pi+i\sin \pi=-1. \end{aligned}$



(b)  Find the solutions to the equation $z^{3}=1$. (That is , find the cube roots of $1$.)

Solution

$\begin{aligned} z^{3}=1&=\cos 0+i\sin 0\\ z=\sqrt[3]{1}&=1^{\frac{1}{3}}\\ &=\cos \frac{0+2k\pi}{3}+i\sin \frac{0+2k\pi}{3},\\ k=0, 1, & 2.\\ &= \cos \frac{2k\pi}{3}+i\sin \frac{2k\pi}{3},\\ k=0, 1, & 2.\\ \end{aligned}$

The cube roots of $1$ are, then

$\begin{aligned} z_0&=\cos 0+i\sin 0=1\\ z_1&=\cos \frac{2\pi}{3}+i\sin \frac{2\pi}{3}=-\frac{1}{2}+i\frac{\sqrt{3}}{2},\\ z_2&=\cos \frac{4\pi}{3}+i\sin \frac{4\pi}{3}=-\frac{1}{2}-i\frac{\sqrt{3}}{2}. \end{aligned}$



(c)  Find the solutions to the equation $z^{4}=1$. (fourth roots of unity)

Solution

$\begin{aligned} z^{4}=1&=\cos 0+i\sin 0\\ z=\sqrt[4]{1}&=1^{\frac{1}{4}}\\ &=\cos \frac{0+2k\pi}{4}+i\sin \frac{0+2k\pi}{4},\\ k=0, 1,& 2, 3.\\ &= \cos \frac{k\pi}{2}+i\sin \frac{k\pi}{2},\\ k=0, 1,& 2, 3.\\ \end{aligned}$

The fourth roots of $1$ are, then

$\begin{aligned} z_0&=\cos 0+i\sin 0=1\\ z_1&=\cos \frac{\pi}{2}+i\sin \frac{\pi}{2}=0+i1=i,\\ z_2&=\cos \pi+i\sin \pi=-1+i0=-1\\ z_3&=\cos \frac{3\pi}{2}+i\sin \frac{3\pi}{2}=-i. \end{aligned}$



(d)  Find the solutions to the equation $z^{5}=1$. (fifth roots of unity)

Solution

$\begin{aligned} z^{5}=1&=\cos 0+i\sin 0\\ z=\sqrt[5]{1}&=1^{\frac{1}{5}}\\ &=\cos \frac{0+2k\pi}{5}+i\sin \frac{0+2k\pi}{5},\\ k=0, 1, & 2, 3, 4.\\ &=\cos \frac{2k\pi}{5}+i\sin \frac{2k\pi}{5},\\ k=0, 1, & 2, 3, 4.\\ \end{aligned}$

The fifth roots of $1$ are, then

$\begin{aligned} z_0&=\cos 0+i\sin 0=1\\ z_1&=\cos \frac{2\pi}{5}+i\sin \frac{2\pi}{5}\\ z_2&=\cos \frac{4\pi}{5}+i\sin \frac{4\pi}{5}\\ z_3&=\cos \frac{6\pi}{5}+i\sin \frac{6\pi}{5},\\ z_4&=\cos \frac{8\pi}{5}+i\sin \frac{8\pi}{5} \end{aligned}$



(e)  Find the solutions to the equation $z^{2}+1=0$. (square roots of $-1$)

Solution

$\begin{aligned} z^{2}+1&=0\\ z^{2}&=-1=\cos \pi+i\sin \pi\\ z&=\sqrt{-1} \\ &=\cos \frac{\pi+2k\pi}{2}+i\sin \frac{\pi+2k\pi}{2},\\ k=0, 1&.\\ & =\cos \frac{2k\pi}{5}+i\sin \frac{2k\pi}{5},\\ k=0, 1&.\\ \end{aligned}$

The square roots of $-1$ are, then

$\begin{aligned} z_0&=\cos \frac{\pi}{2}+i\sin \frac{\pi}{2}=i\quad \text{ and }\\ z_1&=\cos \frac{3\pi}{2}+i\sin \frac{3\pi}{2}=-i. \end{aligned}$



(f)  Find the solutions to the equation $z^{3}+1=0$. (i.e, cube roots of $-1$)

Solution

$\begin{aligned} z^{3}+1&=0\\ z^{3}&=-1=\cos \pi+i\sin \pi\\ z&=\sqrt[3]{-1}\\ &=\cos \frac{\pi+2k\pi}{3}+i\sin \frac{\pi+2k\pi}{3},\\ k=0, 1&, 2. \end{aligned}$

The cube roots of $-1$ are, then

$\begin{aligned} z_0&=\cos \frac{\pi}{3}+i\sin \frac{\pi}{3}=\frac{1}{2}+i\frac{\sqrt{3}}{2}\\ z_1&=\cos \pi+i\sin \pi=-1\\ z_2&=\cos \frac{5\pi}{3}+i\sin \frac{5\pi}{3}=\frac{1}{2}-i\frac{\sqrt{3}}{2}. \end{aligned}$



(g)  Find all solutions to the equation $z^{4}=-256$.

Solution

$\begin{aligned} z^{4}=& -256\\ =& 256(-1)\\ =& 256(\cos \pi+i\sin \pi)\\ =& 256(\cos (\pi+2k\pi)\\ &+i\sin (\pi+2k\pi)\\ z=&(256)^{\frac{1}{4}}(\cos (\pi+2k\pi)\\ &+i\sin (\pi+2k\pi))^{\frac{1}{4}}\\ =& 4\left(\cos \frac{\pi+2k\pi}{4}+i\sin \frac{\pi+2k\pi}{4}\right),\\ k=& 0, 1, 2, 3. \end{aligned}$

All solutions are, then

$\begin{aligned} z_0&=4\left(\cos \frac{\pi}{4}+i\sin \frac{\pi}{4}\right)\\ &=4\left(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}\right)\\ &=2\sqrt{2}+2\sqrt{2}i\\ z_1&=4\left(\cos \frac{3\pi}{4}+i\sin \frac{3\pi}{4}\right)\\ &=4\left(-\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}\right)\\ &=-2\sqrt{2}+2\sqrt{2}i\\ z_2&=4\left(\cos \frac{5\pi}{4}+i\sin \frac{5\pi}{4}\right)\\ &=4\left(-\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}\right)\\ &=-2\sqrt{2}-2\sqrt{2}i\\ z_3&=4\left(\cos \frac{7\pi}{4}+i\sin \frac{7\pi}{4}\right)\\ &=4\left(\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}\right)\\ &=2\sqrt{2}-2\sqrt{2}i\\ \end{aligned}$



Exercise
Try yourself.
  1. Find the sixth roots of unity.
  2. Find the two square roots of $16i$.
  3. Find the solutions to the equation $z^{4}+1=0$
  4. Find the fifth roots of $-1$.
  5. Find the three cube roots of $1+\sqrt{3}i$.



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