Solving $e^{\sin(z)}=1$ in the complex plane

Question

I am trying to solve $e^{\sin(z)}=1$ in the complex plane.

I know that this means that $\sin(z)=2k\pi i$ for some integer $k$. This is equivalent to saying that $$\frac{e^{iz}- e^{-iz}}{2i}=2k \pi i,$$ which means that $$e^{2iz}+4k\pi e^{iz}-1=0.$$ If we let $x=e^{iz}$, then it is a quadratic equation, but my discriminant depends on $k$, so I do not now how to simplify it. Is there an easier way to solve this?

Answer 1

To continue with your line of reasoning (which so far is correct), you need to solve $e^{2iz} + 4 k \pi e^{iz} - 1 = 0$. With $x = e^{iz}$, this becomes the quadratic equation $x^2 + 4kx - 1 = 0$. The discriminant is $\Delta = (4k\pi)^2 + 4 = 4 (4k^2\pi^2 + 1)$, which is always positive ($k$ is real). The two solutions are therefore $x = -2k\pi + \sqrt{4k^2\pi^2+1}$ and $x = -2k - \sqrt{4k^2\pi^2 + 1}$.

• In the first case we want to solve $e^{iz} = -2k\pi + \sqrt{4k^2\pi^2+1}$. Since $4k^2\pi^2+1 = (-2k\pi)^2 + 1> (-2k\pi)^2$, this is a positive real numbers, and so the first batch of solutions is $$z = i\log(-2k\pi + \sqrt{4k^2+1}) + 2 i \pi l, \text{ for some integers } k, l.$$
• In the second case, the equation to solve is $e^{iz} = -2k\pi - \sqrt{4k^2\pi^2+1}$. This is a negative real number, and thus the second batch of solutions is: $$z = i\pi + i\log(2k\pi+\sqrt{4k^2\pi^2+1}) + 2 i \pi l, \text{ for some integers } k, l.$$

And this is the complete set of solutions.

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PS: You can express that a bit more concisely by noticing that $\log(2k\pi + \sqrt{4k^2\pi^2+1}) = \operatorname{argsinh}(2k\pi)$, and so the set of solutions becomes $$e^{\sin z} = 1 \iff z \in \{ i\operatorname{argsinh}(-2k\pi) + 2 i\pi l \mid k,l \in \mathbb{Z} \} \cup \{ i\pi + i\operatorname{argsinh}(2k\pi) + 2 i \pi l \mid k,l \in \mathbb{Z} \}.$$