# $\sqrt{ab}=\sqrt{a}\sqrt{b}$ for complex number $a$ and $b.$

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.

• Solve, $$\large e^{-\frac{xy}{2}}=e^{\frac i2(x+y)}$$ Oct 15, 2015 at 11:10
• If $a,b$ are real, they are also complex, and the equality hold.
– Rick
Oct 15, 2015 at 11:15
• It's not very helpful to think of the square root of a complex number. In the reals it's natural to take the canonical root to be the positive one (where it exists) but of course the negative root is equally a square root. You can't make a continuous choice of canonical square root that works over the whole complex plane: as you know, if you follow a continuous selection all the way around the origin you return with $-1$ times the root you first thought of. Somewhere you have to have a discontinuity.
– HTFB
Oct 15, 2015 at 11:15
• Of course if $x^2 = a$, and $y^2 = b$, then $(xy)^2 = ab$, so in this sense the equality holds everywhere even without picking a single value of $\sqrt{a}$. The product of any square roots of $a$ and $b$ is a square root of $ab$.
– HTFB
Oct 15, 2015 at 11:17
• Then, with the principal branch of the argument, you have the equality if and only if $\lvert \operatorname{Arg} a + \operatorname{Arg} b\rvert < \pi$. Oct 15, 2015 at 11:30

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:

• $a=0$
• $b=0$
• $ab\ne 0$ and $-\pi<\operatorname{Arg}(a)+\operatorname{Arg}(b)\le\pi$
• Why you did't include the case of negative reals....else i got the whole point.... Oct 15, 2015 at 12:24
• you give nice answer please just explane why you did,t include negative reals.. Oct 15, 2015 at 12:26
• Then how to prove that $\sqrt{ab}=\sqrt{a}\sqrt{b}$ does not holds for negative reals $a$ and $b?$ Oct 15, 2015 at 12:31
• $Arg(ab)=Arg(a)+Arg(b)$ if $\frac{-\pi}{2}<Arg(a),Arg(b)\leq\frac{\pi}{2}$ Oct 15, 2015 at 12:38
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.