I recently had this problem on an exam and I was able to find the following solution, but it seemed really ugly and I'm wondering if there is a more elegant approach to this problem.
Find all $x\in \mathbb{R}$ satisfying $e^{xe^{ix}} = e^{ix}$.
My solution was as follows (note that I use $\log$ for the complex valued natural log and $\ln$ for the real valued natural log):
Fix some $x \in \mathbb{R}$ such that $e^{xe^{ix}} = e^{ix}$ holds. First we take the $\log$ of both sides, yielding
\begin{align*}
\log(e^{xe^{ix}}) &= \log(e^{ix})\\
\ln|e^{xe^{ix}}| + i \arg (e^{xe^{ix}}) &= \ln|e^{ix}| + i \arg(e^{ix})\text{ by definition of } \log\\
\ln|e^{x(\cos(x)+i\sin(x))}| + i \arg (e^{xe^{ix}}) &= \ln(1)+i \arg(e^{ix}) \text{ by Euler's formula}\\
\ln(|e^{x\cos(x)}e^{xi\sin(x))}|) + i \arg (e^{xe^{ix}}) &= 0+i \arg(e^{ix})\\
\ln(|e^{x\cos(x)}||e^{xi\sin(x))}|) + i \arg (e^{xe^{ix}}) &=i \arg(e^{ix})\text{ since $|\cdot|$ is multiplicative}\\
\end{align*}
Note that since $x\in \mathbb{R}$, $x\sin(x), x \cos(x) \in \mathbb{R}$ as well, so $|e^{ix\sin(x)}| = 1$ and $e^{x\cos(x)} \in \mathbb{R}$. Thus we have:
\begin{align*}
\ln(|e^{x\cos(x)}|) + i \arg (e^{xe^{ix}}) &=i \arg(e^{ix})\\
x\cos(x) + i \arg (e^{xe^{ix}}) &=i \arg(e^{ix})\\
\end{align*}
Matching real and imaginary parts, we get that
\begin{align*}
x\cos(x) &= 0\\
\arg(e^{xe^{ix}}) &= \arg(e^{ix})
\end{align*}
From the left equality, we get $x \in \{0\} \cup \{\frac{\pi}{2}+n\pi| n \in \mathbb{Z}\}$. Clearly $x = 0$ is a valid solution for the second equality, so it suffices to check which values of $n$ satisfy the second equality. We see that
\begin{align*}
\arg(e^{(\pi/2 +n \pi)e^{i(\pi/2 +n \pi)}}) &= \arg(e^{i(\pi/2 +n \pi)})\\
\arg(e^{(\pi/2 +n \pi)(-1)^ni})&= \frac{\pi}{2} + n \pi \\
(-1)^n(\pi/2 +n \pi) &=\frac{\pi}{2} + n \pi
\end{align*}
Which holds when $n$ is even, so the full solution set is $x \in \{0\}\cup \{\frac{\pi}{2} + 2m \pi| m \in \mathbb{Z}\}$.
Best Answer
$e^{xe^{ix}}=e^{ix} $ if and only if: $\ e^{xe^{ix}-ix}=1\ $.
$e^{xe^{ix}}=e^{ix} $ if and only if: $\ e^{x\cos x} e^{i(x\sin (x)-x)}=1$
$e^{xe^{ix}}=e^{ix} $ if and only if: ($\ x\cos(x)=0 \ $) and ($\ x\sin(x)-x=2k\pi \ $ with $k\in\mathbb Z$)
...