Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm doing some work with complex numbers and I've come across this exercise in the "Polar form" section.


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.



share|cite|improve this question
Take the section name as a hint and start by rewriting $\frac12+\frac{\sqrt3}2 i$ into polar form. – Henning Makholm Dec 6 '12 at 15:08
and then use $(e^{i\theta})^n=e^{i\theta n}$ – Artem Dec 6 '12 at 15:09
Alright, with my calculations, (not messing with ^100) I get $e^{i\pi/3}$ so I should get something like $e^{i\pi/3*100}$ – Rob Dec 6 '12 at 15:18
up vote 2 down vote accepted

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, $(\frac12+i\frac{\sqrt3}2)^{100}=\{(\frac12+i\frac{\sqrt3}2)^3\}^{33}(\frac12+i\frac{\sqrt3}2)=(-1)^{33}(\frac12+i\frac{\sqrt3}2)=-(\frac12+i\frac{\sqrt3}2)$

share|cite|improve this answer

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

share|cite|improve this answer
How did you get $e^{i\pi100/3}$ to $e^{i\pi4/3}$ – Rob Dec 6 '12 at 15:30
@Rob $(e^{i2\pi})^{16}\times e^{i\pi\frac43}$ – Mike Dec 6 '12 at 15:48

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.