Complex: $z^2 = 1+2i$ (find real- and imaginary part of $z$) these are my toughs:
$$z^2 = 1 + 2i \Longrightarrow (x+yi)(x+yi) = 1 + 2i$$
so: $x^2-y^2 = 1$ and $2xy = 2$
then i got that $x = 1/y$ but i cant continue to find the real- and imaginary part of z anymore. Appriciated any help

In general, we have this
Lemma: If $z=a+ib \in \mathbb{C}$, with $a,b\in\mathbb{R}$, then
$$ w = \sqrt{\frac{|z|+a}{2}} + i\epsilon\sqrt{\frac{|z|-a}{2}},$$
where $\epsilon =\pm 1$ according to $b=\epsilon|b|$, satisfies $w^2 = z$.
Proof: Let $w=x+iy$ satisfying $w^2=z$. Then $x^2 - y^2+2xyi =a+bi$.
This equation is equivalent to the system
$$ x^2 - y^2 = a \text{ ; } 2xy = b.$$
Since $w^2 = z$, we have $x^2+y^2 = |z|$ too, and we can conclude that
$$x^2 = \frac{|z|+a}{2} \text{ ; } y^2 = \frac{|z|-a}{2}.$$
Choosing the positive square roots, we can write
$$x = \sqrt{\frac{|z|+a}{2}} \text{ ; }
y = \epsilon\sqrt{\frac{|z|-a}{2}}$$
satisfying $2xy = b$, i.e., $\epsilon = 1$ if $b > 0$ and $\epsilon = -1$ if $b < 0$. 
