Is it correct to solve the equality this way?

Question

$\sinh(z)= i$

$\frac{e^z-e^{−z}}2 = i$

$e^z-e^{−z}= 2 i$

$e^z-2i-e^{−z} = 0$

$e^{z}(e^z-2i-e^{−z}) = 0$

$e^{z}= p$

${p^2}-2ip-1=0\tag{1}$ Quadratic $(1)$ has solutions: $$p_{1,2}= \left(\frac{2i+-\sqrt{(-2i)^2-4\cdot(-1)}}{2}\right) = \frac{2i\pm\sqrt{-4+4}}{2}=i$$

$z \in \operatorname{Ln}{i}$

$z \in {\ln (1) + i (\frac{\pi}{2}+2k\pi)}$.

Is this solution ok? Thank you

Answer 1

It’s fine, and I’m giving you your congratulatory point.

Here’s my way of doing it, just to show that there almost always are more ways to do something than just one. \begin{align} \sinh(z)&=\frac{\sin(iz)}i\quad\text{(must equal $i$ here)}\\ \sin(iz)&=-1\\ iz&=\frac{-\pi}2+2\pi k\\ z&=\frac{i\pi}2+2\pi ki\,, \end{align} just what you got.