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
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 $$ \begin{align} \left(\frac{1+i\sqrt3}{2}\right)^{100} &=e^{i\pi100/3}\\ &=e^{i\pi4/3}\\ &=e^{i\pi}\frac{1+i\sqrt3}{2}\\ &=-\frac{1+i\sqrt3}{2} \end{align} $$
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, $(\frac12+i\frac{\sqrt3}2)^{100}=\{(\frac12+i\frac{\sqrt3}2)^3\}^{33}(\frac12+i\frac{\sqrt3}2)=(-1)^{33}(\frac12+i\frac{\sqrt3}2)=-(\frac12+i\frac{\sqrt3}2)$
