if $z=2+4i$, find $\sqrt{z}$ when $\sqrt{z}=a+bi$ Question
if $z=2+4i$, find $\sqrt{z}$ when $\sqrt{z}=a+bi$
what I have so far
$\sqrt{z}=a+bi$, so I square both sides ->
$z=a^2+2abi-b^2$
we can substitute for z
$z=2+4i=a^2+2abi-b^2$
that means $a^2-b^2=2$ and $2abi=4i$
and ab=2 so then $b=\frac{a}{2}$
from then i dont know how to proceed from there

You have 
$$a^2-b^2=2\quad (1)\\
ab=2\to b=\frac{2}{a}\quad (2)$$
Plug $(2)$ in $(1)$
$$a^2-\frac{4}{a^2}=2\to a^4-2a^2-4=0$$
so
$$a^2=\frac{2\pm\sqrt{20}}{2}=1\pm\sqrt{5}$$
so,
$$a=\pm\sqrt{1+\sqrt{5}}\to b=\pm\frac{2}{\sqrt{1+\sqrt{5}}}=\pm \frac{\sqrt{\sqrt{5}-1}}{2}$$
