Let $z = r(\cos\theta+i\sin\theta)$. In my notes there was this example to calculate the square roots of $i$. What was done was:
$z = r(\cos\theta+i\sin\theta)\\z^2 = r^2(\cos(2\theta)+i\sin(2\theta))\\=\cos(\frac{\pi}{2})+i\sin(\frac{\pi}{2})\\\Leftrightarrow r=1 \ \ \ \ ∧ \ \ \ \ \ 2\theta=\frac{\pi}{2}+2\pi k,\ k\in \mathbb{Z},\ \theta\in[0,2\pi)\\\Leftrightarrow r=1 \ \ \ \ ∧ \ \ \ \ \ \theta=\frac{\pi}{4}+k\pi , \ k=0,1$
Im not entirely understand what they did above, what does the 2 lines below actually tell us? $\Leftrightarrow r=1 \ \ \ \ ∧ \ \ \ \ \ 2\theta=\frac{\pi}{2}+2\pi k,\ k\in \mathbb{Z},\ \theta\in[0,2\pi)\\\Leftrightarrow r=1 \ \ \ \ ∧ \ \ \ \ \ \theta=\frac{\pi}{4}+k\pi , \ k=0,1$
I thought to calculate the square roots of iyou let $z=x+iy$ and work out $x,y$ from $(x+iy)^2=i$?
Also, how did they get that $r^2(\cos(2\theta)+i\sin(2\theta))=\cos(\frac{\pi}{2})+i\sin(\frac{\pi}{2})$?