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.
 A: 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!
A: 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 $\theta = \frac{3\pi}{4} \ \text{or} \ \frac{7\pi}{4}$.
A: 
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
$e^{(\frac {3\pi}4 + k\pi)i} = \cos(\begin{cases}\frac {3\pi}4\\\frac {7\pi}4\end{cases})+\sin(\cos(\begin{cases}\frac {3\pi}4\\\frac {7\pi}4\end{cases}))i =\pm \frac 1{\sqrt 2} \mp \frac 1{\sqrt 2}i$
