Can I say that $1^{\frac{1}{2}} = (e^{2i\pi})^{\frac{1}{2}} =e^{i\pi} = -1 $?

Question

Can I say that $1^{\frac{1}{2}} = (e^{2i\pi})^{\frac{1}{2}} =e^{i\pi} = -1 $ ?

the Answer is obviously no, but why not ?

Answer 1

Actually, you can see it as the solution of a complex equation where the the complex number $z$ and $-z$ are both solutions.