TITLE: When $\cosh (z)=0$?
QUESTION [7 upvotes]: I'm studying complex analysis and I'm wondering about all complex values of $z$ that satisfy the equation:
$$
\cosh(z)=0 \,\, .
$$
Is there a smart way to show all values that vanish with the equation above? If yes, how could I demonstrate this? What are these values? 
Thank you!

REPLY [12 votes]: $$\cosh(z)=0\Longleftrightarrow$$
$$\frac{e^z+e^{-z}}{2}=0\Longleftrightarrow$$
$$e^z+e^{-z}=0\Longleftrightarrow$$
$$e^{-z}\left(1+e^{2z}\right)=0\Longleftrightarrow$$

Since $e^z$ is never zero for any $z\in\mathbb{C}$, no solution exists for $e^{-z}=0$:

$$1+e^{2z}=0\Longleftrightarrow$$
$$e^{2z}=-1\Longleftrightarrow$$
$$2z=i\pi(2n+1)\Longleftrightarrow$$
$$z=\frac{i\pi(2n+1)}{2}$$
With $n\in\mathbb{Z}$