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.



  • 3
    $\begingroup$ Take the section name as a hint and start by rewriting $\frac12+\frac{\sqrt3}2 i$ into polar form. $\endgroup$ – Henning Makholm Dec 6 '12 at 15:08
  • $\begingroup$ and then use $(e^{i\theta})^n=e^{i\theta n}$ $\endgroup$ – Artem Dec 6 '12 at 15:09
  • 1
    $\begingroup$ Alright, with my calculations, (not messing with ^100) I get $e^{i\pi/3}$ so I should get something like $e^{i\pi/3*100}$ $\endgroup$ – Rob Dec 6 '12 at 15:18

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


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

  • $\begingroup$ How did you get $e^{i\pi100/3}$ to $e^{i\pi4/3}$ $\endgroup$ – Rob Dec 6 '12 at 15:30
  • 2
    $\begingroup$ @Rob $(e^{i2\pi})^{16}\times e^{i\pi\frac43}$ $\endgroup$ – Mike Dec 6 '12 at 15:48

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