Question: I have to write $\sum\_{k=0}^{\infty} \cos(k \pi / 6)$ in form: $a+bi$.

I think I should consider $z=\cos(k \pi / 6)+i\sin(k \pi / 6)$ and also use the fact that $\sum \_{k=0}^{\infty}x^n=\frac{1}{1-x}$ if $|x|<1$

But i don't know if thats correct and how can I transform this.

Answer: 
$$\cos\frac{k\pi}6=\begin{cases}1&\;,\;\;k=0,12,24,\ldots=0\pmod{12}\\
\frac{\sqrt3}2&\;,\;\;k=1,11,13,23,\ldots=1,11\pmod{12}\\
\frac12&\;,\;\;k=2,10,14,22,\ldots=2,10\pmod{12}\\etc.\end{cases}$$

You can fill up the lines missing in "*etc.*"...and thus the series cannot possibly converge (why?)
You want divergent summation of real numbers written a complex number. Perhaps the goal of the exercise is to check whether this sum is convergent?

If this is the case, some hints:

* complete your summation with $i\sum\_{k=0}^\infty \sin(k\pi/6) $, you want the real part of this new summation.
* Remember that $cis(kx) = cis(x)^k$
* Observe that the summation can't converge because $cis(x)^n$ diverges