Sqrt(ab) = sqrt(a)sqrt(b)

By Alastair George on Wednesday, February 24, 1999 - 11:08 am :

I would appreciate comments on a question that a colleague and I have been debating.

The problem is to establish whether

\sqrt{ab} = \sqrt{a} \sqrt{b}

Suppose $a$ , $b$ are complex numbers of unit modulus, to make the working simpler, and arguments $a$ , $b$ respectively, with $0 < a$ , $b < 2 π$ .

Then, it would appear that,

$\arg (\sqrt{ab}) = 0.5 (a + b)$ , mod $2 π$

and, similarly,

$\arg (\sqrt{a}) \arg (\sqrt{b}) = 0.5 a + 0.5 b$ , mod $2 π$

The problem, then, is when $a + b = 2 π$

In such a case, the first argument would be returned as zero, whereas the second would be $π$ . However, is it correct to say that for all other combinations of $a + b$ , there is no problem i.e

$\sqrt{ab} = \sqrt{a} \sqrt{b}$ ?
I would be grateful for any comments.

Alastair

By Alex Barnard (Agb21) on Wednesday, February 24, 1999 - 12:29 pm :

You are right that

\sqrt{ab} = \sqrt{a} \sqrt{b}

does not hold sometimes. However it is much more often than just the

a + b = 2 π

case.

For example take $a$ , $b$ both to be the numbers $- i$ (which have arguments $3 π / 2$ . Their product is -1 and its principal square root is $i$ . The principal square root of $- i$ however is $(- 1 + i) / \sqrt{2}$ . When this is multiplied by itself you get $- i$ . So you have problems when $a + b \geq 2 π$ .

The reason you missed this is that your calculations are slightly wrong. They should go:

$\arg (\sqrt{ab}) = 0.5 \arg (ab) = 0.5 (a + b \mod 2 π)$ . Note that the mod is taken BEFORE the 0.5

$\arg (\sqrt{a} \sqrt{b}) = 0.5 a + 0.5 b \mod 2 π$ . Here the mod is taken AFTER the 0.5

Now you can see why there is a problem when $a + b > 2 π$ ... In the second case we never actually use the mod. In the first case we will remove $2 π$ before halving things. Hence our answers will be out by $e^{i π} = - 1$ .

So $\sqrt{ab} = \sqrt{a} \sqrt{b}$ up to the sign.

This is a very common type of behaviour for functions which have what is called a branch point. This is a point where you get a different value back from the function if you walk around it. Imagine going in a circle around 0 and look at what the square root gives you. It will start at 1 and slowly change until when you get back to 1 it will now be -1. This shows that 0 is our branch point. You will find that if you don't circle around 0 then you will get back to exactly where you started from. This shows that 0 is the only branch point.

Write back if you want to know more about functions with branch points...

AlexB.