TITLE: $\sqrt{ab}=\sqrt{a}\sqrt{b}$ for complex number $a$ and $b.$
QUESTION [12 upvotes]: I am trying to find complex numbers $a$ and $b$ for which the equality $\sqrt{ab}=\sqrt{a}\sqrt{b}$ ($\sqrt{z}$ being principal square root)holds. I tried as below 
 $$\sqrt{ab}=\sqrt{a}\sqrt{b}$$ $$\iff$$
$$\sqrt{|ab|}(cos(\frac{Arg(ab)}{2})+isin(\frac{Arg(ab)}{2})=\sqrt{|a|}\sqrt{|b|}(cos(\frac{Arg(a)+Arg(b)}{2})+isin(\frac{Arg(a)+Arg(b)}{2}))$$ $$\iff$$ 
$$Arg(ab)=Arg(a)+Arg(b)$$ Am i right up to this? And can i conclude from above  that $\sqrt{ab}=\sqrt{a}\sqrt{b}$ is true for all positive reals as in this case $Arg(ab)=Arg(a)+Arg(b)=0,$ for one number being positive real and other negative reals as in this case $Arg(ab)=Arg(a)+Arg(b)=\pi?$
And for both number being negative reals then $Arg(ab)=0$ and $Arg(a)+Arg(b)=\pi+\pi=2\pi$. Now i am stuck whether $2\pi=0$ under principal argument or not. Please help me. Thanks in advance.
My question is to find complex numbers under which equality $\sqrt{ab}=\sqrt{a}\sqrt{b}$ holds.

REPLY [3 votes]: By convention, the principal argument function, $\arg(z)$, for nonzero complex numbers $z$ sastisfies the equation $z=|z|e^{i\arg(z)}$ and the inequalities $-\pi\lt\arg(z)\le\pi$.  (It's the two inequalities that confer the modifier "principal.")  The principal square root function is defined as $\sqrt z=\sqrt{|z|}e^{i\arg(z)/2}$ for nonzero $z$ and $\sqrt z=0$ if $z=0$, where $\sqrt{|z|}$ is understood to be the usual postive square root of the positive real number $|z|$ (which itself, recall, is defined as a square root, namely $|z|=\sqrt{x^2+y^2}$).  It's worth noting that if $z$ happens to be a positive real number, then $\arg(z)=0$ and $|z|=z$, so the principal square root function agrees with the usual positive square root function when restricted to the positive real numbers.
With this in mind, the condition you're looking for is actually very simple: For nonzero complex numbers $a$ and $b$, we have 
$$\sqrt{ab}=\sqrt a\sqrt b\iff -\pi\lt\arg(a)+\arg(b)\le\pi$$
As for the question "whether $2\pi=0$ under principal argument or not," the answer is No, $2\pi$ is not equal to $0$.  The whole point of an argument function is to choose a unique angle for each (nonzero) complex number, and the principal argument function chooses the angle $0$, not $2\pi$ for complex numbers lying on the positive real half line.