Elegant solution to $e^{xe^{ix}} = e^{ix}, x \in \mathbb{R}$

Question

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 $x \in \{0\}\cup \{\frac{\pi}{2} + 2m \pi| m \in \mathbb{Z}\}$.

Answer 1

$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 $k\in\mathbb Z$)

...

Answer 2

For $r_1,r_2,\theta_1,\theta_2 \in \mathbb R$ we have $r_1e^{i\theta_1} = r_2e^{i\theta_2}$ if and only if $r_1=r_2$ and $\theta_1-\theta_2 \in 2\pi \mathbb Z := \{2\pi k : k \in \mathbb Z\}$. Thus $e^{-\sin x}e^{i \cos x} = e^{ie^{ix}} = e^{ix}$ implies that $e^{- \sin x} = 1$ and $\cos x - x \in 2\pi \mathbb Z$. Now, from $e^{- \sin x} = 1$ we get $x = k\pi$ for some $k \in \mathbb Z$ and from $$(-1)^k-k\pi = \cos x - x \in 2\pi \mathbb Z$$ we get...

Answer 3

The function $$f(z) : \mathbb C \to \mathbb C \\ f(z) = e^z$$ is periodic with period $2\pi i$, consequently, $g(z) = f(iz)$ is periodic with period $2\pi$. The condition $$e^{ie^{ix}} = e^{ix} \iff g(g(x)) = g(x) \tag{1}$$ therfore requires $$e^{ix} = g(x) = x + 2\pi k \tag{2}$$ for some integer $k$, in the sense that any solution to $(1)$ is a solution to $(2)$ and vice versa. if $x \in \mathbb R$, we must then have $e^{ix} = \cos x + i \sin x = x + 2\pi k$ so $\sin x = 0$, or $x \in \mathbb \pi m$ for some integer $m$. But this would imply $g(\pi m) = (-1)^m$, and $g(g(\pi m)) = e^{\pm i}$, which has a nonzero imaginary part, thus there is no solution.

We can also plot the locus of $$(\Re(g(g(x))-g(x)), \Im(g(g(x)) - g(x)))$$ for $x \in [0, 2\pi)$:

This shows that this function has no zeros for $x \in \mathbb R$.

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Since the question was edited to fix an error, the solution is radically different.

Let $x = \theta + 2\pi k$ for some $\theta \in [0,2\pi)$ and $k \in \mathbb Z$. Then $g(x) = e^{ix} = \cos \theta + i \sin \theta$, and $$\begin{align} h(x) &= e^{x g(x)} - g(x) \\ &= e^{x \cos \theta} e^{i x \sin \theta} - \cos \theta - i \sin \theta \\ &= e^{x \cos \theta} \cos (x \sin \theta) - \cos \theta + i \left( e^{x \cos \theta} \sin (x \sin \theta) - \sin \theta \right). \end{align}$$ As we require both real and imaginary parts to be zero, this implies $$\cos \theta = e^{x \cos \theta} \cos (x \sin \theta) \\ \sin \theta = e^{x \cos \theta} \sin (x \sin \theta).$$ It follows from taking the sum of the squares that $$e^{2x \cos \theta} = 1.$$ Since the exponent is real-valued, we conclude $2x \cos \theta = 0$, which is true if and only if $x = 0$ or $\theta \in \{\frac{\pi}{2}, \frac{3\pi}{2}\}$. In the first case, we obtain $$h(0) = 0,$$ and in the second, we find $x = 3\pi/2$ to be extraneous. Just to check, we compute $$g(\pi/2 + 2\pi k) = e^{i \pi/2} = i,$$ hence $$h(\pi/2 + 2\pi k) = e^{(\pi/2 + 2\pi k)i} - i = e^{i \pi/2} - i = 0,$$ so the complete solution set is $$x \in \{0, \pi/2 + 2\pi k\}$$ for $k \in \mathbb Z$.