Why is it that $ \ln(1) = 0 $ but $ e^{(i2\pi)}=1$?

Question

Why is it that $ e^{(i2\pi)} = 1 $ but $ \ln(1) = 0 $? In other words, is $2\pi i = 0$? I know that $e^0 = 1$, but should it also equal $e^{2\pi i}$?

Answer 1

It doesn't. $$\ln{(1)}=2\pi ki\text{ for all }k\in\mathbb{Z}$$ You are just assuming that the principal valued natural logarithm is the only branch.

Answer 2

Just because two exponentials ($e^{2\pi i}$ and $e^0$ in this case) are equal, that doesn't mean that the exponents ($2\pi i$ and $0$ in this case) are equal.

When we write $\ln(1)$, and expect a single answer, then we have to make a choice: which exponent do we actually return? Usually, the exponent $0$ is chosen. But that doesn't mean that it's the only possible answer, and it doesn't even really mean that it is better than any other answer. It's just the most conventional.