$z^2 = -1$ Why what i do is wrong? I want to calculate $z^2 = -1$ i think i should get  to $z = i$.
$-1 = -1+0i = 1(\cos(2\pi k)+i\sin(2\pi k)), k \in Z$
$z^2 = r^2(\cos(2 \theta) + i\sin(2\theta))$
So i get: 
$$r^2(\cos(2 \theta) + i\sin(2\theta)) = 1(\cos(2\pi k)+i\sin(2\pi k))$$
Namely: 
$r = 1, 2 \theta = 2 \pi k \Rightarrow \theta = \pi k$
So for $k = 0$ i get $z_0 = 1$ for $k=1: z_1 = -1$
What i do wrong?

$$(a+b\,i)^2=a^2-b^2+2ab\,i=-1\iff a^2-b^2=-1,2ab=0.$$
The only way is by $a=0,b^2=1$, hence $\pm i$.

Your mistake was in $2k\pi$ instead of $(2k+1)\pi$.
