Why isn't the identity $\sqrt{ab}$ = $\sqrt{a} \cdot \sqrt{b}$ always true?

Question

If we take $a=b=-1$ then the L.H.S. is $1$ but the R.H.S. is $-1$.

1. Is this identity not applicable for complex numbers?

2. How to prove this and prove that this is not applicable for some complex numbers?

Answer 1

The root function behaves differently for complex numbers, because:

• Over the reals, we only take the positive number, so $\sqrt{9} = 3$, but not $\sqrt{9} = -3$
• Over the complex numbers, we take all possible roots, so $\sqrt{-9} = 3i$, but also $\sqrt{-9} = -3i$. That is the fundamental difference. In fact when we say we use the complex square root, we have $\sqrt{9} = 3$ and $\sqrt{9} = -3$.

It is indeed confusing that we use the same symbol for a different function.

Then I want to say that $i^2=1$, but $i$ is not $\sqrt{-1}$. We have that $\sqrt{-1}=\pm i$.

Answer 2

It is not true for all complex numbers, no.

You've given precisely a counter-example to the claim that $\sqrt{ab}=\sqrt{a}\cdot \sqrt{b}$ holds.