How to convert from complex number to polar coordinates (solve for r and theta).

Question

How can I convert $ze^z$ to polar coordinates and what are the answers for ($r$ and $\theta$)?
I've been trying to solve the question for so long, but I unfortunately didn't succeed.

Answer 1

Let $x=ze^z$, $z=re^{(i\theta)}=r(cos\theta+isin\theta)$.
So, $x=re^{i\theta}e^{r(cos\theta+isin\theta)}=re^{i\theta}e^{rcos\theta+irsin\theta}$
Therefore, $x=re^{rcos\theta}e^{i(\theta+sin\theta)}$
On solving, $$x=re^{rcos\theta}(cos(\theta+sin\theta)+isin(\theta+sin\theta))$$
Or, $$x=|z|e^{Re(z)}(cos(\theta+sin\theta)+isin(\theta+sin\theta))$$
Or, $$x=|z|e^{Re(z)}\Bigg(cos\bigg(\theta+\frac{Im(z)}{r}\bigg)+isin\bigg(\theta+\frac{Im(z)}{r}\bigg)\Bigg)$$ The previous three equations are same, consider whichever suffices your usage.

Answer 2

Let $z = p e^{i\psi}$, where $|z| = p$ and $arg(z) = \psi$. This implies that,$$ \begin{align} z e^z &= p e^{i \psi} e^{p e^{i \psi}} \\ &= p e^{i \psi}e^{pcos \psi + i p sin \psi} \\ &= p e^{i \psi}e^{pcos \psi} e^{ i p sin \psi} \\&= p e^{pcos \psi} e^{i (\psi + p sin \psi)} \end{align}$$ Thus, we have written $z e^z$ in the polar form where $r = p e^{pcos \psi}$ and $\theta = \psi + p sin \psi$.