What it says in the title. I've got the proof mostly down, but I must have made an error somewhere in my ''calculations'', as I'm getting $x = \frac{n\pi} 2$ for the real part of the complex number $z = x + \mathrm i y$.
My work is as follows:
\begin{align} &&\frac{\mathrm e^{\mathrm i z} - \mathrm e^{-\mathrm i z}} {2\mathrm i} &= 0\\ \iff &&\mathrm e^{\mathrm i z} - \mathrm e^{-\mathrm i z} &= 0\\ \iff &&\mathrm e^{\mathrm i z} &= \mathrm e^{-\mathrm i z}\\ \iff &&\mathrm e^{\mathrm i (x + \mathrm i y)} &= \mathrm e^{-\mathrm i (x + \mathrm i y)}\\ \iff &&\mathrm e^{-y + \mathrm i x} &= \mathrm e^{y - \mathrm i x}\\ \iff &&\mathrm e^{-y}\Big( \cos x + \mathrm i \sin x \Big) &= \mathrm e^{y}\Big( \cos x - \mathrm i \sin x \Big)\\ \iff &&\frac{ \cos x + \mathrm i \sin x }{ \cos x - \mathrm i \sin x } &= \mathrm e^{2y}\\ \iff &&\frac{ (\cos x + \mathrm i \sin x)^2 }{ (\cos x - \mathrm i \sin x)(\cos x + \mathrm i \sin x) } &= \mathrm e^{2y}\\ \iff &&\frac{ (\cos x + \mathrm i \sin x)^2 }{ \cos^2 x + \sin^2 x } &= \mathrm e^{2y}\\ \iff &&\frac{ (\cos x + \mathrm i \sin x)^2 }{ 1 } &= \mathrm e^{2y}\\ \iff &&(\cos x + \mathrm i \sin x)^2 &= \mathrm e^{2y}\\ \iff &&\cos^2 x - \sin^2 x + 2\mathrm i \cos x \sin x &= \mathrm e^{2y}, \end{align}
Now the imaginary part of this should be zero for the equation to hold, meaning \begin{equation} 2\sin x \cos x =\sin 2x = 0 \iff x = \frac {n\pi} 2\ldots \end{equation}
Not what I wanted to see. So, where did I make a mistake? I can't seem to spot it, to no great surprise.
Suppose $x = \frac{n\pi}2$. If $n$ is odd, we have $\cos x = 0$ and $\sin x = 1$. In this case, the equation $$\cos^2 x - \sin^2 x + 2i \cos x \sin x = e^{2y}$$ simplifies to $$-1 = e^{2y},$$ which is impossible since the right-hand side is positive.
Therefore $n$ is even.