Why is the second equality wrong?

Question

Here's a "proof" of $e^x=1$ for all $x$: $$\exp(x)=\exp\left(i2π⋅\frac{x}{i2π}\right)=\bigl(\exp(i2π)\bigr)^{x/(i2π)}=1^{x/(i2π)}=1$$ Why is the second equality wrong?

Answer 1

This argument is false because $h(u,v)=u^v$ is not a single-valued function in the complex numbers. In particular, $1^v\neq 1$ for some values of $1^v$.

If you choose some single-valued branch of $u^v$, then it is not true that $\exp(xv)=\exp(x)^v$ in general.

The wonderful, magical thing about $\exp(x)$ is that it is a function - it is well-defined for all complex $x$. But in general exponentiation, we are not so lucky. If $a$ is a complex number, then we usually define $a^x$ to be $\exp(x\ln a)$, but $\ln a$ is not well-defined - there are infinitely many different possible values for $\ln a$.