Finding the (complex) solutions $z$ to $z^4 = w$ 
What are possible solutions to $$z^4 = w$$ where $z \in  \mathbb{C} $ and $w \in \mathbb{R}$?

Here is my attempt:
Write both numbers in polar form: $r^4(\cos 4 \theta + i\sin 4 \theta) = w(\cos 0 + i \sin 0) $.
Working with the conventions of $r \geq 0$ and $0 \leq \theta < 2\pi$ I get the following:
$r^4 = w \iff r = w^\frac{1}{4}$ 
and 
$4\theta = n2\pi$ where $n \in \mathbb{Z}$.
$\Rightarrow \theta = \frac{n}{2}\pi$
Thus we get four possible solutions:
$z = w^\frac{1}{4}(\cos 0 + i \sin 0) = w^\frac{1}{4}$ 
$z = w^\frac{1}{4}(\cos \frac{\pi}{2} + i \sin \frac{\pi}{2})= iw^\frac{1}{4}$ 
$z = w^\frac{1}{4}(\cos \pi + i \sin \pi) = -w^\frac{1}{4}$ 
$z = w^\frac{1}{4}(\cos \frac{3\pi}{2} + i \sin \frac{3\pi}{2}) = -iw^\frac{1}{4}$ 
Is this legit, something missing or very much not rigorous?

If $w < 0$, then the equation $r = w^{1 / 4}$ is not even sensible, as any fourth root of $w$ is nonreal (and anyway in this case there is not a preferred choice among these) but $r$ is a real variable. The fix is to write any $w < 0$ in polar form as $$w = |w|e^{\pi i} = |w|(\cos \pi + i \sin \pi).$$ Then, regardless of the sign of $w$ we have $r = |w|^{1 / 4}$, and when $w < 0$ our equation in $\theta$ becomes $4 \theta = (2n + 1) \pi$.
