Q:

Why is $1=\sqrt{(-1)(-1)}=\sqrt{i^2\cdot i^2}=\sqrt{i^4}=\sqrt{\left(e^{i\frac{\pi}{2}}\right)^4}=\sqrt{e^{i2\pi}}=e^{i\pi}=-1$ wrong?

Why is $1=\sqrt{(-1)(-1)}=\sqrt{i^2\cdot
 i^2}=\sqrt{i^4}=\sqrt{\left(e^{i\frac{\pi}{2}}\right)^4}=\sqrt{e^{i2\pi}}=e^{i\pi}=-1$
  wrong?

my friend showed me this "proof" - what's the first step that is not allowed? Isn't  $1=\sqrt{(-1)(-1)}$ wrong?

A:

The only incorrect statement is that
$$\sqrt{e^{i2\pi}}=e^{i\pi}$$
Because we have that
$$\sqrt{x^2}=|x|$$
for real $x$. So we would have the answer as
$$\sqrt{e^{i2\pi}}=|e^{i\pi}|=|-1|=1$$

A:

There is nothing wrong with the equality $1=\sqrt{(-1)(-1)}$. The problem lies in the equality $\sqrt{e^{2\pi i}}=e^{\pi i}$, because $e^{2\pi i}=1$, and therefore $\sqrt{e^{2\pi i}}=1\neq e^{\pi i}$.

