How do I find all complex numbers that are solutions of $z^3 +8i=0$? My workbook has answers in the standard form $a+bi$. I would assume that to solve this, I would expand the complex number $z$ into trigonometric form to deal with that exponent. This is what I have so far using De Moivre's Theorem:
$(r^3(\cos(3\theta)+i\sin(3\theta)) +8i=0$. 
However, I do not know where to go from here. Thank you!
 A: $$z^3=-8i\iff
\begin{cases}
|z|^3=|-8i|=8\\
\arg{z^3}=3\arg z=-\dfrac{\pi}{2}+2k\pi
\end{cases}$$
$\iff$
$$
\begin{cases}
|z|=2\\
\arg{z}=-\dfrac{\pi}{6}+\dfrac{2\pi}{3}k, \quad k\in\{0,1,2\}
\end{cases}$$
Can you take it from here?
A: Well, $-8i=2^3\exp\frac{-i\pi}{2}$, so one cube root is $2\exp\frac{-i\pi}{6}$. You can get the rest yourself.
A: I took a rather naive approach.
\begin{align}
z^3+8i&=0\\
(z)^3-(2i)^3&=0\\
(z-2i)(z^2+2iz+4i^2)&=0\\
(z-2i)(z^2+2iz-4)&=0
\end{align}
\begin{align}
z&=\frac{-(2i)\pm\sqrt{(2i)^2-4(1)(-4)}}{2}\\
z&=\frac{-2i\pm\sqrt{-4+16}}{2}\\
z&=\frac{-2i\pm\sqrt{12}}{2}\\
z&=-i\pm\sqrt3
\end{align}
The solution set is therefore $z=2i$, $z=\sqrt3-i$, and $z=-\sqrt3-i$.
A: You found that $r^3 \cos(3\theta)+i r^3 \sin(3\theta)=-8i$. We have that $|r^3 \cos(3\theta)+i r^3 \sin(3\theta)|=|-8i|$ and therefore $r^3=8$ and $r=2$.
Now we have $\cos(3\theta)+i \sin(3\theta)=-i$. Since real part on the right side is $0$, it must be $0$ on the left too. Since imaginary part on the right side is $-1$, it must be $-1$ on the left side too. We have that $\cos(3\theta)=0$ and $\sin(3\theta)=-1$. This gives us: $$3\theta=2k\pi+\frac{3\pi}{2}$$
$$\theta=\frac{2k\pi}{3}+\frac{\pi}{2}$$.
If it looks like there are infinitely many solutions, it's an illusion.
Since $z$ depends only on the $\sin$ and $\cos$ of the angle $\theta$, we will have the same $z$ for $\theta$ and $\theta+2k\pi$. There are 3 solutions for every $\theta \in (0,2\pi)$ that satisfies the condition.
A: For questions like these the Euler form works the best.
Substitute $z=re^{i\theta}$ to get $r^3e^{3i\theta+2n\pi}=-8i$
Now take cube root on both sides then different values of $n$ would give you different values of $z$.
A: You have $r^3(\cos(3\theta)+i\sin(3\theta))=c$ with $\theta\in\Bbb R,r\in[0,\infty)$, therefore you have to go $$\begin{cases}r^3=\lvert c\rvert\\ \cos(3\theta)=\frac1{\lvert c\rvert}\Re c\\ \sin(3\theta)=\frac1{\lvert c\rvert}\Im c\end{cases}$$
A: $(r^3(\cos(3\theta)+i\sin(3\theta)) +8i=0$
So continue:
$(r^3(\cos(3\theta)+i\sin(3\theta)) = -8i= 8(0 + i(-1))$
$(r^3(\cos(3\theta)+i\sin(3\theta)) = 8((\cos \frac {3\pi}2+2k\pi) + i\sin(\frac{3\pi}2+2k\pi))$
So $r^3 = 8$ and $3\theta = \frac {3\pi}2 + 2k\pi$
So $r =2$ and $\theta = \frac \pi 2 + \frac 23k\pi$ 
So there are three solutions:
1)if $k =0$ then $2(\cos \frac \pi 2 + i\sin \frac \pi 2) = 2i$.
Check.   $(2i)^3 = 2^3i^2 = 8*i^2 *i = -8i$.  That's good.
2) if $k = 1$ then $2(\cos (\frac \pi 2+\frac {2\pi} 3) + i \sin (\frac \pi 2+\frac {2\pi} 3))= 2(\cos \frac {7\pi}6) + i\sin \frac {7\pi}6) = -\sqrt 3 - i$.
Check:  $(-\sqrt 3-i)^3 = -[(\sqrt 3)^3 + 3(\sqrt 3)^2i + 3(\sqrt 3)i^2 + i^3] = $
$-[3\sqrt 3 + 9i - 3\sqrt 3 - i] = 8i$.  It works.
and if $k = 2$ then 3) $2(\cos (\frac \pi 2+\frac {4\pi} 3) + i \sin (\frac \pi 2+\frac {4\pi} 3))= 2(\cos \frac {11\pi}6) + i\sin \frac {11\pi}6) = \sqrt 3 - i$.
Check:  $(\sqrt 3-i)^3 = (-\sqrt 3)^3 + 3(-\sqrt 3)^2i + 3(-\sqrt 3)i^2 + i^3] = $
$-3\sqrt 3 - 9i +3\sqrt 3 + i = -8i$.  It works.
A: You could just eyeball it and see that $2^3=8, i^3=-i$ 
so $(2i)^3=-8i$ so the solutions are $2i, 2i\omega, $ and $2i\omega^2$, where $\omega$ is a complex cube root of $1$.
But to continue your way,  $r^3(\cos(3\theta)+i\sin(3\theta))=-8i \implies r^3=8$ and $\sin(3\theta)=-1,$ 
i.e., $3\theta=\dfrac{3\pi}2+n2\pi$, so $\theta=\dfrac{\pi}2+\dfrac{2n}3\pi,$
so solutions are $2\exp\dfrac{i\pi}2=2i, 2\exp\dfrac{i7\pi}6=-\sqrt3-i,$ and $2\exp\dfrac{i11\pi}6=\sqrt3-i,$
