Suppose
$a,b\in \mathbb{Z}$.
Is it true
$\sqrt{a}\sqrt{b}=\sqrt{ab}$.
If so, then $\sqrt{-1}\sqrt{-1}=\sqrt{(-1)(-1)}=\sqrt{1}=1$
But we know $\sqrt{-1}=i$ and so $i^2=-1.$
Finally we get $i^2=-1=1.$
Which is not true.
What is the logic behind it?
Thank you in advance.
The statement is true for positive, real $a, b$, but in general it does not hold for complex numbers. This can be seen as a consequence of the fact that we don't have a natural choice of inverse for $f(z) = z^2$ on $\mathbb{C}$. For certain choices of inverses (i.e., choices of square root functions), the above statement holds for complex numbers $a, b$ whose arguments satisfy certain inequalities, but when first learning about complex numbers, it's perhaps safest to treat the identity as though it only holds for positive reals.