Find all the solutions for $e^z=e^{iz}$

Question

so iam trying to solve this

$e^z=e^{iz} $

so, since z = x+iy

$ e^{x+iy} = e^{-y+ix} $

so should i take the Log for the two sides ? or what should i do ?!!

I really want the help!

Thanks

Answer 1

Hint:

$$e^{z(1-i)}=1=e^{2m\pi i}$$ where $m$ is any integer

Answer 2

If $z,w\in\mathbb C$, then $e^z=e^w$ if and only if there is an integer $k$ such that $z=w+2k\pi i$. So, when do we have $z-iz=2k\pi i$ for some integer $k$?

Answer 3

Notice that $$|e^z|=|e^{iz}|$$therefore $$e^x=e^{-y}$$which leads to $$x=-y$$therefore $$e^z=e^{iz}=e^{x+ix}=e^{x-ix}$$which reduce to $$e^{ix}=e^{-ix}$$hence the solutions are$$x=-y=k\pi$$and $$z=k\pi (1-i)\quad,\quad k\in \Bbb Z$$