# Pulling a negative out of a square root

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1. Jan 18, 2016

### Mr Davis 97

The following is invalid, since the operation is not defined when $a, b < 0$: $\sqrt{-1}\sqrt{-1} = \sqrt{(-1)(-1)} = \sqrt{(-1)^2} = \sqrt{1} = 1$. This is not correct, because $ii = -1$. This shows that $\sqrt{a}\sqrt{b} = \sqrt{ab}$ is invalid when $a, b< 0$.

However, say we have $\sqrt{-5}$. In order to simplify this, we do the following: $\sqrt{-5} = \sqrt{(-1)(5)} = \sqrt{-1}\sqrt{5} = i\sqrt{5}$. Why is this a valid manipulation given the previous statement that $\sqrt{a}\sqrt{b} = \sqrt{ab}$ is invalid when $a, b< 0$?

2. Jan 18, 2016

### phinds

You are applying a criterion where a and b are both less than zero to a situation where only one of them is less than zero. I don't see why you would expect it to apply.

3. Jan 18, 2016

### BvU

In my math book (1971 ) the range of the complex square root function is $-{1\over 2} \pi \lt \arg {\bf z} \le {1\over 2} \pi$.

With the range of $\phi = \arg {\bf z}$ : $- \pi \lt \phi \le \pi$, and using euler $\ {\bf z} \equiv |{\bf z}| e^{i\phi}$ we can now define $$\ \sqrt {\bf z} \equiv\; \sqrt{|{\bf z}|} \; e^{i {\phi\over 2} }$$
With this definition of the complex square root you can see that $\sqrt {\bf \alpha\beta} = \sqrt {\bf \alpha} \sqrt {\bf \beta}$ only holds if $\ |\arg {\bf \alpha} + \arg {\bf \beta} | < \pi$

Check out the euler formula, de moivre theorem.