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. 
 A: $$z^2=-81i\Longleftrightarrow$$
$$z^2=|-81i|e^{\arg(-81i)i}\Longleftrightarrow$$
$$z^2=81e^{-\frac{\pi i}{2}}\Longleftrightarrow$$
$$z=\left(81e^{\left(2\pi k-\frac{\pi}{2}\right)i}\right)^{\frac{1}{2}}\Longleftrightarrow$$
$$z=9e^{\frac{1}{2}\left(2\pi k-\frac{\pi}{2}\right)i}$$
With $k\in\mathbb{Z}$ and $k:0-1$
So, the solutions are:
$$z_0=9e^{\frac{1}{2}\left(2\pi\cdot0-\frac{\pi}{2}\right)i}=9e^{-\frac{\pi i}{4}}$$
$$z_0=9e^{\frac{1}{2}\left(2\pi\cdot1-\frac{\pi}{2}\right)i}=9e^{\frac{3\pi i}{4}}$$
Now, notice that:
$$\color{red}{\frac{\pi}{2}<\arg\left[9e^{\frac{3\pi i}{4}}\right]<\pi\Longleftrightarrow\frac{\pi}{2}<\frac{3\pi}{4}<\pi}$$
So, your right answer is:
$$9e^{\frac{3\pi i}{4}}=9\cos\left(\frac{3\pi}{4}\right)+9\sin\left(\frac{3\pi}{4}\right)i=-\frac{9\sqrt{2}}{2}+\frac{9\sqrt{2}}{2}i$$
A: I think you have understood why the modulus is $81$. 
Now, since $r^2(cos(2θ)+isin(2θ)) = -81i$
 clearly, $[cos(2θ)]+[sin(2θ)]i = -i = (0) + (-1)i$. 
Since the real and imaginary parts of the complex number are independent, you can equate them to get the answer. 
So, you need to solve the equations $$cos(2\theta) = 0,\,\ sin(2\theta)=-1$$  
A: 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
$$
\sqrt[n]{r}\left(
   \cos\left(\frac{\alpha}{n}+\frac{2k\pi}{n}\right)
 +i\sin\left(\frac{\alpha}{n}+\frac{2k\pi}{n}\right)
\right),
\qquad k=0,1,\dots,n-1
$$
A: I would highly recommend using De Moivre's formula and graphing.
Start by getting some graphing paper and label your $x$-axis as real values from $-100$ to $100$ and label your $y$-axis as imaginary values from $-100i$ to $100i$.  (go by $10$'s for the intervals.)
Place the complex number you are trying to square root on the graph.
Measure the angle it makes with the positive $x$-axis, the axis-line that points right.  The angle measure for any real positive values is $0\deg$, naturally, and you can go from there?  (Note: we measure our angles counterclockwise)
Now, all you have to do to square root this number is simply to 'half' the angle and square root the magnitude.
$$\text{magnitude}=|a+bi|=\sqrt{a^2+b^2}$$
In your case, $|-81i|=81$, so the square root of that is $9$.
Since we can see the angle measure is $270\deg$, half of that is $135\deg$.  To calculate the exact value of that, the trigonometric equation is DeMoivre's Formula:
$$(a+bi)^n=|a+bi|^n\cdot(\cos(n\theta)+i\sin(n\theta))$$
For our case, $\sqrt{-81i}=(-81i)^{0.5}=81^{0.5}\cdot(\cos(0.5\cdot270)+i\sin(0.5\cdot270))=9(\cos(135)+i\sin(135))=9(-\frac{\sqrt2}{2}+i\frac{\sqrt2}2)$
Lastly, note that $270\deg=270+360\deg$ because the extra $360$ is just another spin around the circle, which ends up back at $270\deg$.  However, when you half this 'new' angle, you get a different result, which notably explains why we sometimes have $x^2=a\implies x=\pm\sqrt a$ instead of $x=\sqrt a$.
If you want more insight on how DeMoivre's formula comes about, I recommend graphing $(1+i)^n$ for $n=0,1,2,3,\dots$ and measuring the angle measure and magnitude.  (You can calculate $(1+i)^3=(1+i)(1+i)(1+i)$, for example.  Just foiling)
