Q:

Sum of complex number

I tried to solve the following Sum of complex number:

$\sum_{k=0}^{32} (\frac{1}{2}+i\frac{\sqrt{3}}{2})^k$

I proceeded like this  :

$\sum_{k=0}^{32} (\frac{1}{2}+i\frac{\sqrt{3}}{2})^k$= $\sum_{k=0}^{32}e^{i\frac{2k\pi}{3}}$=$\frac{1-e^{i\frac{2*33\pi}{3}}}{1-e^{i\frac{2\pi}{3}}}$= $\frac{1-e^{{10i\pi}}}{1-e^{i\frac{2\pi}{3}}}$=$0$

The only problem is that $|e^{i\frac{2k\pi}{3}}|$=1. and not less then 1.

A:

What's the problem? The equality$$\sum_{k=0}^Nz^k=\frac{1-z^{N+1}}{1-z}$$holds whenever $z\neq1$. So, your approach is correct. But note that$$\frac12+\frac{\sqrt3}2i=\exp\left(\frac{\pi i}3\right).$$Therefore, the answer is$$\frac{1-\exp\left(11\pi i\right)}{1-\exp\left(\frac{\pi i}3\right)}=\frac2{1-\exp\left(\frac{\pi i}3\right)}=1+\sqrt3\,i.$$

