Why is $\arg(i\cosh x)=\frac{\pi}{2}$?

Question

I was told $\arg(i\cosh (x))=\frac{\pi}{2}$ and $\arg(\cosh (x))=0$ but I can't figure out why. Could someone explain it to me?

Answer 1

Here I assume $x\in\mathbb{R}$, then $cosh(x)>0$ is a real number.

Since argument of $z$ is the angle from the positive real axis to the vector representing $z$, thus $i\cosh(x)$ has $\frac{\pi}{2}$ as argument and $\cosh(x)$ has $0$ as argument. More precisely, here "argument" means principal argument.