Simplifying $z^2+i=0$ I need to simplify $z^2+i=0$ and find all solutions for $z$. I have seen that the solutions to $z=\sqrt{i}=\left(\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}\right)$ and $\left(-\frac{1}{\sqrt{2}}-i\frac{1}{\sqrt{2}}\right)$. I was hoping to find a similar solution for $z=\sqrt{-i}\,$ but my attempt gives me $z=\pm i^{\frac{3}{2}}$
$$z=re^{i\theta} \,\,\& \,\, e^{i\pi}=-1 $$
then,
$$(re^{i\theta})^2=-i\\r^2e^{i2\theta}=ie^{i\pi+k(2\pi)}$$ where $k\in\mathbb{Z}$.
So, we have $\begin{cases} r^2=i \,\,\,\therefore r=\sqrt{i}\\ \theta=k\pi 
\end{cases}$
Then, $$z_k=\sqrt{i} \, e^{i\left(\frac{\pi}{2}+k\pi\right)}$$
$$z_0=\sqrt{i}\left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right)=i^\frac{3}{2}\\ 
z_1=\sqrt{i}\left(\cos(\frac{\pi}{2}+1)+i(\sin\frac{\pi}{2}+1)\right)=-i^\frac{3}{2}$$
I realized that this is literally the same as just solving $z=\sqrt{-i}=i\sqrt{i}=i^\frac{3}{2}$, however, I was hoping to find a solution of the form $x+iy$. I am not sure how to go about this problem a  different way.


I was hoping to find a similar solution for z=−i−−√ but my attempt gives me z=±i32

I don't see why.
$z = x+yi$ and $z^2 = (x^2 -y^2) + 2xyi = -i$ so $x^2-y^2 = 1$ and $2xy = -1$ so $x^2 =y^2$ and $x = \pm |y|$ but $2xy =-1$ is negative os $x = -y$ and so $2xy=-1\implies -2y^2 = 1\implies y =\pm \frac 1{\sqrt 2}$ and $x = \mp \frac 1{\sqrt 2}$. and $z = \pm \frac 1{\sqrt 2} \mp \frac 1{\sqrt 2} i$.
Which shouldn't surprise us as if $(\pm \frac 1{\sqrt 2} \pm \frac 1{\sqrt 2} i)^2 = i$ then $i(\pm \frac 1{\sqrt 2} \pm \frac 1{\sqrt 2} i) = \pm \frac 1{\sqrt 2}i \mp \frac 1{\sqrt 2} =\mp \frac 1{\sqrt 2} \pm \frac 1{\sqrt 2}$ when squared should be $i^2*i = -i$.
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Using polar coordidates $-i = 0 + (-1)i = \cos \frac {3\pi}2 +\sin(\frac {3\pi}2)i = e^{(\frac {3\pi}2 + 2k\pi)i}$ and so the square roots of $-i$ are
$e^{(\frac {3\pi}4 + k\pi)i} = \cos(\begin{cases}\frac {3\pi}4\\\frac {7\pi}4\end{cases})+\sin(\cos(\begin{cases}\frac {3\pi}4\\\frac {7\pi}4\end{cases}))i =\pm \frac 1{\sqrt 2} \mp \frac 1{\sqrt 2}i$
