Polar form of Complex numbers I'm doing some work with complex numbers and I've come across this exercise in the "Polar form" section.
$$(1/2+i(\sqrt{3}/2))^{100}$$
Of course this exercise is manageable with the help of Pascal's triangle and numerous hours of calculating, but I assume there's a much simpler solution.
The result should be: $$-(1/2)-i(\sqrt{3}/2)$$
I'm appreciative with any possible help.
Thanks,
Bobby
 A: The Euler's formula says that
$$
e^{i\phi}=\cos(\phi)+i\sin(\phi)
$$
So we get that
$$
\frac{1+i\sqrt3}{2}=e^{i\pi/3}
$$
So raising both sides to the power $100$, remembering $e^{i2\pi}=1$, yields
$$
\begin{align}
\left(\frac{1+i\sqrt3}{2}\right)^{100}
&=e^{i\pi100/3}\\
&=e^{i\pi4/3}\\
&=e^{i\pi}\frac{1+i\sqrt3}{2}\\
&=-\frac{1+i\sqrt3}{2}
\end{align}
$$
A: Let $\frac12+i\frac{\sqrt3}2=R(\cos\theta+i\sin\theta)$  where $R\ge 0$
Equating the real & the imaginary part, $R\cos\theta=\frac12$ and $\frac{\sqrt3}2=R\sin\theta$
Squaring & adding we get, $R^2=1\implies R=1$
On division, $\tan\theta=\sqrt 3$ so that $\theta=\frac \pi3$ as both $\sin\theta,\cos\theta>0$
So, $\frac12+i\frac{\sqrt3}2=\cos\frac \pi3+i\sin\frac \pi3$
Using de Moivre's formula/identity, $(\frac12+i\frac{\sqrt3}2)^3=(\cos\frac \pi3+i\sin\frac \pi3)^3=\cos\pi+i\sin\pi=-1$
Hence, $(\frac12+i\frac{\sqrt3}2)^{100}=\{(\frac12+i\frac{\sqrt3}2)^3\}^{33}(\frac12+i\frac{\sqrt3}2)=(-1)^{33}(\frac12+i\frac{\sqrt3}2)=-(\frac12+i\frac{\sqrt3}2)$
