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?

Assume $z\in\mathbb{C}$ and $w\in\mathbb{R}$:
$$z^4=w\Longleftrightarrow$$
$$z^4=|w|e^{\arg(w)i}\Longleftrightarrow$$
$$z^4=we^{0i}\Longleftrightarrow$$
$$z=\left(we^{2\pi ki}\right)^{\frac{1}{4}}\Longleftrightarrow$$
$$z=\sqrt[4]{w}e^{\frac{2\pi ki}{4}}\Longleftrightarrow$$
$$z=\sqrt[4]{w}e^{\frac{\pi ki}{2}}$$
With $k\in\mathbb{Z}$ and $k:0-3$

So the solutions are:
$$z_0=\sqrt[4]{w}e^{\frac{\pi\cdot 0i}{2}}=\sqrt[4]{w}e^{0}=\sqrt[4]{w}$$
$$z_1=\sqrt[4]{w}e^{\frac{\pi\cdot 1i}{2}}=\sqrt[4]{w}e^{\frac{\pi i}{2}}=i\sqrt[4]{w}$$
$$z_2=\sqrt[4]{w}e^{\frac{\pi\cdot 2i}{2}}=\sqrt[4]{w}e^{\pi i}=-\sqrt[4]{w}$$
$$z_3=\sqrt[4]{w}e^{\frac{\pi\cdot 3i}{2}}=\sqrt[4]{w}e^{\frac{3\pi i}{2}}=-i\sqrt[4]{w}$$
So we can conclude that your solutions are:
$$z=\pm\sqrt[4]{w} \space\space\vee\space\space z=\pm i\sqrt[4]{w}$$
