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
 A: 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 $b < 0$. 
A: You have $x^2 - y^2 = 1$ and $x = \frac{1}{y}$. Substitute the latter into the former to get $$\frac{1}{y^2} - y^2 = 1 \implies 1 - y^4 = y^2 \implies y^4 + y^2 - 1 =0.$$
You can solve the quadratic $u^2 + u -1 = 0$ giving solution $u = \frac{1}{2}(-1 \pm \sqrt{5})$. So you need to solve $y^2 = \frac{1}{2}(-1 + \sqrt{5})$ (since $y$ is real) which is now straightforward. 
A: An alternative approach would be to use DeMoivre's Theorem. 
 $1+2i=\sqrt{5}e^{i\arctan{2}}$ so $z=\sqrt[4]{5}e^{i\arctan 2/2}$ or $z=\sqrt[4]{5} e^{(i\arctan 2 +2\pi i)/2}$  If $\tan \theta =2, \tan \frac{\theta}{2} = \frac{2}{1+\sqrt{5}}$ giving $$z=\sqrt[4]{5}\frac{1+\sqrt{5}}{\sqrt{10+2\sqrt{5}}}+i\sqrt[4]{5}\frac{2}{\sqrt{10+2\sqrt{5}}}$$  The other root being $-z$.
A: \begin{align}
z^2 = 1 + 2i & = |1+2i|(\cos\alpha+i\sin\alpha) \text{ where } \tan \alpha = \frac 2 1 \\[10pt]
& = \sqrt{1^2+2^2} (\cos\alpha+i\sin\alpha)  = \sqrt 5 (\cos\alpha+i\sin\alpha)
\end{align}
Therefore
$$
z = \pm\left( \sqrt{\sqrt 5} \right)\left( \cos\frac\alpha2 + i \sin\frac\alpha 2 \right).
$$
