Find all $z$ such that $e^z = i$

Question

So I have to find all solutions of $e^z=i$. Instead of the usual method (write $z = x+yi$ and compare real and imaginary parts), I wrote $e^z = e^{iz/i}$, and therefore

$$\cos\left(\frac{z}{i}\right) + i\sin\left(\frac{z}{i}\right) = \cos\left(2k\pi+\frac{\pi}{2}\right) + i\sin\left(2k\pi+\frac{\pi}{2}\right),$$

which gave me the correct solution $z = \left(2k\pi+\frac{\pi}{2}\right)i$. Is there something illegal with this? My main concern are the factors $\cos\left(\frac{z}{i}\right)$ and $\sin\left(\frac{z}{i}\right)$, given that their argument $\frac{z}{i}$ is now potentially a true complex number, which could mess up the value of $\sin$ and $\cos$. Thanks!

Answer 1

Your approach is correct. But I would suggest to write instead $$i=0+i\cdot 1=\cos(\pi/2+2k\pi)+i\sin(\pi/2+2k\pi)=e^{i(\pi/2+2k\pi)}$$

Answer 2

Mixing algebraic and exponential form is faster:

Set $z=x+iy$. You obtain the equation $$\mathrm e^x\mathrm e^{iy}=i=\mathrm e^{i\tfrac\pi 2}\iff\begin{cases}\mathrm e^x=1\iff x=0,\\[1ex]y\equiv\dfrac\pi 2\mod 2\pi. \end{cases}$$