Is $0$ a imaginary number?

Question

I was using Euler's formula $ e^{ix} = \cos(x) + i \sin(x)$ with the case when $ x = 2\pi$. $$ e^{i2\pi} = \cos(2\pi) + i\sin(2\pi) = 1 + 0i $$ So I end up with the identity $ e^{i2\pi} = 1$. Now, if I apply $\ln$ it goes $$ \ln e^{i2\pi}=\ln1$$ $$ i2\pi = \ln1$$ How $\ln1=0$, Should be that $i2\pi=0$ its true, or not? The first implication I found its that you can divide any imaginary number by zero. $$ {ai\over 0} = {ai\over i2\pi} = {a\over 2\pi}$$ What don you think?

Answer 1

You are assuming that there is a function $\ln\colon\Bbb C\setminus\{0\}\longrightarrow\Bbb C$ such that, if $z\in\Bbb C$, $\ln\bigl(e^z\bigr)=z$. But there is no such function.