If $a,b \in \mathbb R$, prove that the equation $z^3=a+bi$ has at least one complex solution.
It seem to me pretty obvious that this equation has 3 complex roots and they are $\sqrt[3]{a+bi}$, $(-1)^{\frac{2}{3}} \sqrt[3]{a+bi}$ and $(-1)^{\frac{4}{3}} \sqrt[3]{a+bi}$ and they're all complex. So that's the whole solution? I would love to have strict proof of this.
I assume that the goal of the problem is to give an elementary proof of the fact that any complex number has at least one cubic root. In your solution, you write $\sqrt[3]{a+bi}$, but isn't this what we want to prove?
Maybe you can say: every real number has at least one real cubic root (Bolzano's Theorem) and every complex number with modulo $1$ is of the form $e^{\alpha i}$, hence has also at least one complex cubic root, namely $e^{\frac{\alpha}{3}i}$. Now any $z\in \mathbb{C}$ can be written as $re^{\alpha i}$ for some $r, \alpha \in \mathbb{R}$, and therefore we can find at least one complex cubic root of $z$ by finding one for $r$, one for $e^{\alpha i}$ and then multiplying them.