Multiplication of the square root of complex numbers

Question

$$ \sqrt{-21+20i}.\sqrt{-21-20i}=\pm(2+5i).\pm(2-5i)=\pm29 $$ But why it is not $$ \sqrt{-21+20i}.\sqrt{-21-20i}=\sqrt{(-21+20i)(-21-20i)}=\sqrt{|z|^2}\\ =\sqrt{441+400}=\sqrt{841}=29 $$

Does this has something to do with $\sqrt{-a}\sqrt{-b}=-\sqrt{ab}\neq\sqrt{ab},\;a,b\in\mathbb{R_+}$ ?

Where does all these rules coming from ?

Answer 1

In the context of complex numbers, it is not a good idea to write $\sqrt z$, since every complex number (other than $0$) has two square roots. So, unless you specify which square root you have in mind, the expression $\sqrt z$ is ambiguous. And all you can say about the square roots of a product is that each of them can be obtaind multiplying some square root of the first factor by some square root of the second one.

Answer 2

Square root of any non-zero complex number is two valued function since $(-z)^2=z^2$. In the case of the positive real numbers (and only in this case) both roots are real numbers. The positive of them is termed principal square root and usually is denoted as $\sqrt x$. To avoid misunderstanding it is therefore advantageous to use $x^\frac12$ instead of $\sqrt x$.

And $841^\frac12=\pm29$.

Answer 3

By definition, $\sqrt(x) = y$ means that $y^2 = x$, so both $y$ and $-y$ would work. However, in case of real numbers, square root is conventionally defined as a non-negative root.

In your case $\sqrt(29^2) = \sqrt((-29)^2) = 29$, so both roots are valid.