<s>[INST] How do I find the solution to the equation $z^2=-81i$? This question is from the Powers of Complex Numbers, Precalculus section of KhanAcademy
Find the solution to the following equation whose argument is between $90°$ and $180°$
$$z^2=-81i$$
What I understand thus far:
I am going to set $r$ and $\theta$ to be the modulus and argument of $z$, respectively.
Therefore, $z^{ 2 }=r^{ 2 }[cos(2\cdot \theta )+isin(2\cdot \theta )]$
Now, I can understand how the modulus is $81$, but I do not understand how it was determined that the argument is $270°$ plus any multiple of $360°$. I am quite confused at this point and a hint in the right direction would be the best thing to help me figure out the solution to this problem and ones like it that I will encounter in the future. 
 [/INST] Write $-81i$ in trigonometric form:
$$
-81i=81\cdot(-i)=81\left(\cos\frac{3\pi}{2}+i\sin\frac{3\pi}{2}\right)
$$
so by De Moivre its square roots are
$$
9\left(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}\right)
$$
and
$$
9\left(\cos\left(\frac{3\pi}{4}+\pi\right)+
i\sin\left(\frac{3\pi}{4}+\pi\right)\right)
=
9\left(\cos\frac{7\pi}{4}+i\sin\frac{7\pi}{4}\right)
$$
Since $\pi/2<3\pi/4<\pi$, the first one is what you're looking for. Thus the answer is
$$
9\left(-\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}\right)
$$
If you use degrees, $\frac{3\pi}{2}=270^\circ$ and $\frac{3\pi}{4}=135^\circ$.

In general, if $z=r(\cos\alpha+i\sin\alpha)$, the $n$-th roots of $z$ are
$$
\sqrt[n]{r}\left(
   \cos\left(\frac{\alpha}{n}+\frac{2k\pi}{n}\right)
 +i\sin\left(\frac{\alpha}{n}+\frac{2k\pi}{n}\right)
\right),
\qquad k=0,1,\dots,n-1
$$
</s>