Find all the values in $\Bbb C$ of $sech^{-1}(i)$
My solution is First recall that $sech^{-1}(z)=\log(\frac{1}{z}+\sqrt{(\frac{1}{z^2}-1)})$, Then evaluating at $z=i$ we get that $sech^{-1}(i)=\log(\frac{1}{i}+\sqrt{-2}) = \log(-i+2i)=\log(i)$ Now recall that $\log(i)=Log(|z|)+i(\arg(z)+2k\pi)$ where $k\in \Bbb Z$ where $\arg(z)=\theta$ and $\theta \in (-\pi,\pi]$ then substituting into the formula we get that $\log(i)=i(\frac{\pi}{2}+2k\pi)$, is this correct?, and if so when i draw an Argand diagram would it just be points on the imaginary axis at every $(\frac{\pi}{2}+2k\pi)$ for $k \in \Bbb Z$ ?
The question seems to be looking for all values of $z$ so that $$ \frac2{e^z+e^{-z}}=i $$ which means $$ e^{2z}+2ie^z+1=0\implies e^z=i\left(-1\pm\sqrt2\right) $$ That is, $$ z=\pm\left(\log\left(\sqrt2+1\right)+\left(2k-\frac12\right)\pi i\right) $$