Power when exponent is i

Question

Suppose we compute $z=3^i$. $$3^i=(3*1)^i=3^i*(e^{2\pi ni})^i=3^i*e^{-2\pi n}\Rightarrow 3^i*(1-e^{-2\pi n})=0\Rightarrow 3^i=0,\textrm{ if }n\neq 0$$

So is $z=0$ a solution too ?

Answer 1

The correct way: Write $$3^{i}=e^{\ln 3^i}= e^{i \ln 3}= \cos(\ln 3)+i \sin (\ln 3).$$

In your step $3^{i}$ remains as it is, you have made no progress.