$z^2 = -1$ 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?

It should be
$-1 = -1+0i = 1(\cos(\pi(2 k+1))+i\sin(2\pi(2 k+1))), k \in Z$
$z^2 = r^2(\cos(2 \theta) + i\sin(2\theta))$
So  $r^2(\cos(2 \theta) + i\sin(2\theta)) = 1(\cos(\pi(2 k+1))+i\sin(2\pi (2k+1)))$
Namely, $r = 1, 2 \theta = \pi(2 k+1)\Rightarrow \theta = \pi (k+\frac12)$.
So for $k = 0$ you get $z_0 = i$, and, for $k=1, z_1 = -i$.
