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's the solution to $z^2 = i$. Give $z^2 + i = 0$ a try after looking through this.
You have the right idea in using $ z = re^{i\theta} $. First, note that we can choose $r = 1$ because $|re^{i\theta}| = |r|$, and $|i| = 1$. Then, substitute:
$$ z^2 = i$$
$$ (e^{i\theta})^2 = i $$
$$ e^{i(2\theta)} = i $$
$$ \cos(2\theta) + i\sin(2\theta) = i . $$
Comparing the real parts of the left and right-hand sides, we have
$$ \cos(2\theta) = 0 $$
$$ \Rightarrow 2\theta = \frac{\pi}{2} + \pi k $$
$$ \Rightarrow \theta = \frac{\pi}{4} + \frac{\pi}{2}k $$
Now comparing the imaginary parts, we have
$$ \sin(2\theta) = 1 $$
$$ \Rightarrow 2\theta = \frac{\pi}{2} + 2\pi k $$
$$ \Rightarrow \theta = \frac{\pi}{4} + \pi k $$
The two solutions have an intersection of $ \theta = \frac{\pi}{4} + \pi k $. In one rotation from $\theta = 0$ to $\theta = 2\pi$, that gives us two unique solutions:
$$ \theta = \frac{\pi}{4} \ \text{or} \ \theta = \frac{5\pi}{4} .$$
Substituting these back into $z = e^{i\theta} = \cos(\theta) + i\sin(\theta) $, we get the solutions you pointed out:
$$ \theta = \frac{\pi}{4} \Rightarrow z = \frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i , $$
$$ \theta = \frac{5\pi}{4} \Rightarrow z = -\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i . $$
Now, for $z^2 = -i$, the solution should be a rotation of $\frac{\pi}{2}$ from the solution we got here, i.e., you should get to $\theta = \frac{3\pi}{4} \ \text{or} \ \frac{7\pi}{4}$.
