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.

Here it is another solution for the sake of curiosity.
Let $z = x + yi$, where $x,y \in \mathbb{R}$. Then we have the following equation:
\begin{align*}
z^{2} + i = 0 & \Longleftrightarrow (x+yi)^{2} = x^{2} - y^{2} + 2xyi = -i\\\\
& \Longleftrightarrow
\begin{cases}
x^{2} - y^{2} = 0\\\\
2xy = -1
\end{cases}\\\\
 & \Longleftrightarrow
\begin{cases}
x = -y\\\\
2y^{2} = 1\\
\end{cases}\\\\
& \Longleftrightarrow
\begin{cases}
x = -\dfrac{1}{\sqrt{2}}\\\\
y = +\dfrac{1}{\sqrt{2}}
\end{cases};
\begin{cases}
x = +\dfrac{1}{\sqrt{2}}\\\\
y = -\dfrac{1}{\sqrt{2}}
\end{cases}
\end{align*}
Hopefully this helps!
