solving a basic complex equation but using de Moivres theorem I have a question which should be super super easy!
If I was to solve $z^2 = 1+i$ how would I do this using de-moivres theorem?
I have the answer here in front of me so I don't want the answer, I just dont understand the method very well!
Any help would be appreciated! I haven't had much experience with complex numbers and have just started a complex analysis course.
Many thanks

$$\forall w=x+iy\in\Bbb C:$$
$$w=|w|e^{i\phi}=|w|(\cos\phi+i\sin\phi)\;,\;\;\phi=\begin{cases}\arctan\frac{y}{x}+2k\pi &,\;\;y\neq 0\\{}\\2k\pi\end{cases}\;\;,\;\;\;k\in\Bbb Z$$
In our case:
$$w=1+i\Longrightarrow |w|=\sqrt 2\,\,,\,\,\arctan\frac{1}{1}=\frac{\pi}{4}+2k\pi\Longrightarrow$$
$$z^2=1+i=\sqrt 2\,e^{\frac{\pi i}{4}\left(1+8k\right)}\;\;,\;\;k=0,1\;\Longrightarrow z=\sqrt[4]2\, e^{\frac{\pi i}{8}(1+8k)}\;,\;\;k=0,1$$
Why do we restrict ourselves only to the above values of $\,k\,$ ? Because any other integer value will give one of these two different ones (on the trigonometric circle, say) ! 
Thus, the solutions are 
$$k=0:\;\;\;\;z_0:=\sqrt[4]2\,e^{\frac{\pi i}{8}}=\sqrt[4]2\left(\cos\frac{\pi}{8}+i\sin\frac{\pi}{8}\right)\\k=1:\;\;\;z_1:=\sqrt[4] 2\,e^{\frac{9\pi i }{8}}=\sqrt[4] 2\left(\cos\frac{9\pi}{8}+i\sin\frac{9\pi}{8}\right)$$
By the way, it is a nice exercise to show that $\,z_0=-z_1\,$...
