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.
Best Answer
Your equivalences are close, but not quite correct, and not well-justified. The crucial missing condition is: $$a,b,ab\in\Bbb C\setminus\{t\in\Bbb R:t\le 0\}\tag{$\star$}$$
Note that $\sqrt{|ab|}=\sqrt{|a||b|}=\sqrt{|a|}\sqrt{|b|}$ by properties of positive real numbers. Thus, $\sqrt{ab}=\sqrt{a}\sqrt{b}$ if and only if both $(\star)$ and $$\operatorname{cis}\frac{\operatorname{Arg}(ab)}2=\operatorname{cis}\frac{\operatorname{Arg}(a)+\operatorname{Arg}(b)}2\tag{1}$$ hold. (Here, I use the abbreviation $\operatorname{cis}\theta:=\cos\theta+i\cdot\sin\theta$.) From this, we cannot directly conclude that $\operatorname{Arg}(ab)=\operatorname{Arg}(a)+\operatorname{Arg}(b),$ because the cosine and sine functions are periodic! Thus, $(1)$ holds if and only if $$\frac{\operatorname{Arg}(ab)}2=\frac{\operatorname{Arg}(a)+\operatorname{Arg}(b)}2+2\pi k\tag{2}$$ for some integer $k,$ which holds if and only if $$\operatorname{Arg}(ab)=\operatorname{Arg}(a)+\operatorname{Arg}(b)+4\pi k\tag{3}$$ for some integer $k.$
Now, by definition of principal argument, the left-hand side of $(3)$ is a number in the interval $(-\pi,\pi],$ while the right-hand side of $(3)$ can (and should) be shown to be an element of $(-2\pi+4\pi k,2\pi+4\pi k].$ For any integer $k\ne0,$ these two intervals are disjoint, and so we conclude that $k=0,$ whence $(3)$ holds if and only if $$\operatorname{Arg}(ab)=\operatorname{Arg}(a)+\operatorname{Arg}(b).\tag{4}$$
Now, on the one hand, it is easy to show that if $\lvert\operatorname{Arg}(a)+\operatorname{Arg}(b)\rvert>\pi,$ then $(4)$ does not hold, and if $\lvert\operatorname{Arg}(a)+\operatorname{Arg}(b)\rvert=\pi,$ then $(\star)$ does not hold. Hence, if $\lvert\operatorname{Arg}(a)+\operatorname{Arg}(b)\rvert\ge\pi,$ then $\sqrt{ab}=\sqrt{a}\sqrt{b}$ does not hold. By contrapositive, $$\sqrt{ab}=\sqrt{a}\sqrt{b}\implies\lvert\operatorname{Arg}(a)+\operatorname{Arg}(b)\rvert<\pi.$$
On the other hand, it is fairly straightforward to prove that $$\lvert\operatorname{Arg}(a)+\operatorname{Arg}(b)\rvert<\pi\implies\sqrt{ab}=\sqrt{a}\sqrt{b},$$ giving us a nice equivalence.
Added: You may be wondering why we bother with the condition $(\star),$ in the first place. After all, we could easily say that $\sqrt0=0,$ and use $\sqrt{z}=\sqrt{|z|}\operatorname{cis}\frac\pi2=i\sqrt{|z|}$ for $z$ negative real, so why don't we just do that? Well, there are a number of reasons not to, actually.
One very nice thing about the function $\sqrt t$ on the positive reals is that it is continuous, and even differentiable! We'd really like the complex version to have the same property, which is impossible to do if we try to extend it to the whole plane. Indeed, consider the function $z(\theta)=\operatorname{cis}\theta$, and note that $\lim_{\theta\to\pi}z(\theta)=\lim_{\theta\to-\pi}z(\theta)=-1.$ However, we find that $$\lim_{\theta\searrow-\pi}\sqrt{z(\theta)}=-i\ne i=\lim_{\theta\nearrow\pi},$$ and so we cannot continuously define the function at $-1$ (or any other negative real number), much less differentiably!
Now, if we relax our requirements very slightly, we can continuously (though not differentiably) extend the function by saying $\sqrt0=0,$ but we can't do any better if we require continuity.
If we drop continuity, though, why not just let $\sqrt{-t}=i\sqrt t$ for positive real $t$? Well, that invites the question of what $i$ is, in the first place. It is a square root of $-1.$ But there is no substantial reason why we should give $i$ any preference over its opposite when choosing a value for $\sqrt{-1}.$
Of course, we certainly can extend it to be defined everywhere as you suggest. If we do so, then $\sqrt{ab}=\sqrt{a}\sqrt{b}$ holds if and only if one of the following holds: