$i^2 = (-i)^2$ apparently contradiction

Question

I got this question in a class and can't figure out the answer:

$i^2 = (-i)^2 = -1$

$(e^{πi(2n + 1/2)})^2 = (e^{πi(2m - 1/2)})^2$

$e^{πi(4n + 1)} = e^{πi(4m - 1)}$

$\ln(e^{πi(4n + 1)}) = \ln(e^{πi(4m - 1)})$

$πi(4n + 1) = πi(4m - 1)$

$m = n + 1/2$

Which isn't possible for m and n both being integers. But I can't figure out what exactly is wrong with what is written here.

Answer 1

You're making an assumption that's correct in the reals, but not in the complex numbers - you're assuming that if $e^a = e^b$, then $a = b$. Notice that $e^{2\pi i} = \cos(2\pi) + i\sin(2\pi) = 1 + 0i = 1 = e^0$, but $2\pi i \neq 0$. The thing is that when you're working with complex numbers, $e^a = e^b$ just tells you that $a = b + 2\pi in$ for some $n \in \mathbb{Z}$.