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 $z^{4}=-256$.
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