How to solve $z^6+i=0$ I'm trying to solve $z^6+i=0$. I would have say that it's equivalent to $$z^6=-i\iff |z|^6e^{i6\arg(z)}=e^{i\frac{3\pi}{2}}\iff|z|^6=e^{i\left(\frac{3\pi}{2}-6\arg(z)\right)}$$
But I'm not able to conclude that $z\in\left\{e^{i\frac{3+4k}{12}}\mid k=0,1,...,5\right\}$
 A: Not quite, I recommend going about this problem a different way. $$z^6=-i\\ \implies z^6 =e^{i\frac{(4k+3)\pi}{2}}\\ \implies  z=\left(e^{i\frac{(4k+3)\pi}{2}}\right)^{\frac{1}{6}} \\ \implies z= e^{i\frac{(4k+3)\pi}{12}}$$ where we have six different solutions for $k \in \{0,1,2,3,4,5 \}$ which allows you to conclude that $$z \in \left\{e^{i\frac{(4k+3)\pi}{12}} \lvert \space k=0,1,2,3,4,5\right\}$$
A: First, represent $-\rm i$ in exponential form
$$
z^6 = -{\rm i} = {\rm e}^{- \frac{{\rm i}\pi}2}
$$
Now you can multiply it by $1 = {\rm e}^{2\pi \rm i}$, actually you can do it $k$ times:
$$
z^6 = {\rm e}^{2{\rm i} \pi k - \frac{{\rm i}\pi}2}, k =0,1,\ldots
$$
now let us consider the $1/6$-th power of the left an of the right sides of the equation
$$
z = {\rm e}^{\left(2{\rm i} \pi k - \frac{{\rm i}\pi}2\right)/6} 
= {\rm e}^{\frac{4 k - 1}{12}{\rm i}\pi} 
= {\rm e}^{\frac{4 k + 3}{12}{\rm i}\pi} 
$$
A: You're close:
$$
z^6 =-i= e^{i\big( \frac{3\pi}{2} + 2\pi k\big)}
$$
for $k \in \Bbb{Z}$. With $z = re^{i \theta}$ you find that $r^6 = 1$ implying that $r=1$ and $\theta = \frac{3\pi}{12} + \frac{\pi k}{3}$ for $k \in \Bbb{Z}$. Can you go from here..?
A: Factor it as $$(z^2+i)(z^4+iz^2-1)=0$$ and solve these separately instead. If you don't wanna use the quadratic formula for the second, you can substitute $y=iz^2$ and complete the square. 
Why is everyone going the hard way anyway?
A: You already got that $-i=\mathrm{e}^{3\pi i/2}$. Let $z=r\,\mathrm{e}^{i\vartheta}$.
Then
$$
z^6=r^6\,\mathrm{e}^{6\vartheta i}=\mathrm{e}^{3\pi i/2}.
$$
Hence $r=1$, and
$$
6\vartheta=3\pi/2+2k\pi,\quad k\in\mathbb Z
$$
or
$$
\vartheta=\frac{\pi}{4}+\frac{k\pi}{3},
$$
and
$$
z=\mathrm{e}^{\vartheta i}=\mathrm{e}^{i\frac{\pi}{4}+i\frac{k\pi}{3}}.
$$
