Prove that there are always two square roots of a non zero complex number

How would I do this?

- Oct 23rd 2009, 09:42 PMdifferentiateProve that there are always two square roots of a non zero complex number
Prove that there are always two square roots of a non zero complex number

How would I do this? - Oct 23rd 2009, 09:48 PMchisigma
- Oct 23rd 2009, 09:59 PMBruno J.
Well that's one way to show it, but I'd compare it to killing a fly with a nuclear bomb! (Smirk)

Every complex number can be written as (polar form). Then and are the two square roots of ( denotes the usual, positive square root of the real number ). I'll let you show that they are distinct when . That there cannot be any others is a consequence of the fact that polynomials over a field (such as ) cannot have more roots than their degree. - Oct 24th 2009, 04:59 AMtonio
- Oct 24th 2009, 06:22 AMHallsofIvy
Yet another way. By DeMoivres' theorem, the kth roots of the complex number are of the form for integer j.

Here, k= 2 so the square roots are given by .

If j is any even number, say j= 2n, so that

If j is odd, say j= 2n+ 1, so .

The only two square roots are and .