Why $\ln(1)\neq 2\pi ik$

Question

Given that $e^{2\pi ik}=1$ for all $k \in \mathbb{Z}$, why isn't $\ln{e^{2\pi ik}}=2\pi ik$? On the other hand $\ln(1)=0$. What am I missing here?

Answer 1

By (my) definition of $\log$, for any $x > 0$, $\log x = \int_1^x \frac 1 t dt$. So $\log 1 = \int_1^1 \frac 1 t dt = 0$.

By some other definition, for any $x > 0$, $\log x$ is the real number $y$ such that $\exp y = x$ (because$\exp : \Bbb R \to \Bbb R_{>0}$ is bijective). The only real number $y$ such that $\exp y = 1$ is $0$ so $\log 1 = 0$.

On the other hand, you have just discovered that the exponential function on complex numbers ( $\exp : \Bbb C \to \Bbb C^*$) is not injective, so there can't be a corresponding inverse function $\log : \Bbb C^* \to \Bbb C$.