Why is $\sqrt{xy}=\sqrt{x}\sqrt{y}$ also true when $x=-1$, making it $i$?

Question

A : If: $$\sqrt{xy}=\sqrt{x}\sqrt{y}$$ only when $x,y>0$,

B : Then why can I do this:

$$\sqrt{-4}=\sqrt{4\times-1}=\sqrt{4}\sqrt{-1}=2i$$

which violates A since $y<0$

C : But why can I not do this?

$$\sqrt{4}=\sqrt{-1\times-1\times4}=\sqrt{-1}\sqrt{-1}\sqrt{4}=i\cdot i\cdot2=-2$$

Which follows the same reasoning as B.

Your first statement $A$ is wrong..$\sqrt{xy}=\sqrt x\sqrt y$ holds iff at least one of $x$ and $y$ is positive. — Empty, Jul 16 '15 at 07:19
Because the condition $x,y>0$ doesn't mean that it can NEVER be true if $y<0$, it just means that there's no reason the expect it to be true. So you B case is just an example that 'just happens' to work. remember 'false$\Rightarrow$ true' holds. — user2520938, Jul 16 '15 at 07:20
$\sqrt{xy}=\sqrt{x}\sqrt{y}$ is also for $x>0,y<0$ or $x<0,y>0$ — Chiranjeev_Kumar, Jul 16 '15 at 07:20

score 2 · Accepted Answer · edited Apr 13 '17 at 12:20

We have $\sqrt{xy}=\sqrt{x}\sqrt{y}$ if and only if $-\pi < \arg x + \arg y \le \pi$, where $\sqrt x$ denotes the principal square root and $\arg x$ denotes the principal argument. See this answer for details.

If $x$ and $y$ are positive numbers, then $\arg x + \arg y = 0 + 0 = 0$, so the identity holds.

In your case B, $\arg(4) + \arg(-1) = 0 + \pi = \pi$, so the identity still holds.

In your case C, $\arg(-1) + \arg(-1) = \pi + \pi = 2\pi$, so the identity fails.

Why is $\sqrt{xy}=\sqrt{x}\sqrt{y}$ also true when $x=-1$, making it $i$?

1 Answers1