Solve $e^z + e^i = 0$

Question

Solve $e^z + e^i = 0$

Normally I would set

$e^z = e^i$ and then use $z = i$

However in this case I have $e^z = -e^i$ which confuses me. Also I'm not sure if it's possible to go to $z = i$ since $Log$ isn't defined for all complex numbers.

Answer 1

$$e^z=-e^i=(\cos\pi+i\sin\pi)e^i=e^{i\pi}\cdot e^i$$

$$=e^{(2n+1)i\pi}e^i=e^{(2n+1)i\pi+i}$$

$$\implies z=(2n+1)i\pi+i$$ where $n$ is any integer

Answer 2

$$-e^i=(-1)e^i=e^{(2k+1)\pi i}e^i=e^{[1+(2k+1)\pi]i}$$

Answer 3

$$e^z+e^i=0\iff e^z=-e^i\iff e^{z-i}=-1\iff z-i=i\pi+2ik\pi,\;k\in\Bbb Z$$ and we solve the last equation easily.