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?

Question

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?

Answer 1

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 $\sqrt{e^{2\pi i}}=1\neq e^{\pi i}$.

Answer 2

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 $$\sqrt{e^{i2\pi}}=|e^{i\pi}|=|-1|=1$$

Answer 3

Square root is only a single valued function if you restrict yourself to positive real numbers. This is something that people are at least casually aware of when they solve quadratic equations: if you want to solve $x^2=4$, you "take the square root of both sides", but you get two roots, $x=\pm 2$.

When you're not working in a situation where there is a single valued square root function, you need to take the multi-valued aspect more seriously. This means that we no longer have that $\sqrt{z^2}=z$, but rather $\sqrt{z^2}=\{z,-z\}$, and therefore $z\in \sqrt{z^2}$. Another way to patch this is that, if $z$ is real, $|z|=\sqrt{z^2}$. In either case, the identity $\sqrt{z^2}=z$ is only true when $z$ is a positive real number, and cannot be used more generally.

Others have said that the problem is the equality $\sqrt{e^{2\pi i}}=e^{\pi i}$, and while it is true that you only have $\sqrt{e^{2\pi i}}\in \{\pm e^{\pi i}\}$ or $\sqrt{e^{2\pi i}}=|e^{i\pi}|$, it is also true that the problem started with the statement that $1=\sqrt{(-1)(-1)}$, as that is only true is you if you are talking about the single-valued positive square root function, and while you never took the square root of a non-positive number, the manipulations you do make it clear that you want to consider $\sqrt{z}$ more generally.

Answer 4

The first step is wrong already.

In $\mathbb C$ we have $\sqrt1=\{1,-1\}$.

Actually, in $\mathbb R$ we have $\sqrt1=1.$