# Square root of negative complex exponential

1. Feb 7, 2012

### amalak

1. The problem statement, all variables and given/known data

Solve $\sqrt{-e^{(i2\pi)/3}}$

2. Relevant equations

3. The attempt at a solution

I seem to be missing something simple, as I take:

$\sqrt{-1}$ = i

then,

$e^{(1/2)*(i2\pi)/3}$

which comes out as: $ie^{i\pi/3}$

however, the solution is:

$-ie^{i\pi/3}$, and I can't seem to see where that negative is coming from. Any direction would be great, thanks!

2. Feb 7, 2012

### susskind_leon

The rule sqrt(a*b)=sqrt(a)*sqrt(b) isn't valid for complex numbers.
Consider sqrt((-1)*(-1))=sqrt(-1)*sqrt(-1)=i*i=(-1) which is incorrect, obviously.

3. Feb 7, 2012

### vela

Staff Emeritus
You want to find the principal square root, which is defined to be $\sqrt{z} = re^{i\theta/2}$ when you have $z = re^{i\theta}$ with $-\pi < \theta \le \pi$. You just need to find the correct $\theta$ for your case.

4. Feb 7, 2012

### amalak

So in this case, the correct θ is not $\pi/3$? Since, we're trying to solve $\theta/2$, wouldn't $(2\pi/3)/2 = \pi/3$ be in our desired range? Also, how would I handle the -1 in this case if that rule is not valid for complex numbers?\

I think I solved this:

$-e^{(i2\pi)/3} = e^{-i\pi/3}$, then $\sqrt{e^{-i\pi/3}} = e^{-i\pi/6}$, as desired. This required De Moivre's formula, is there another way to go about this solution?

Last edited: Feb 7, 2012
5. Feb 7, 2012

### vela

Staff Emeritus
$$e^{-i\pi/6} = (-i^2)e^{-i\pi/6} = -i(ie^{-i\pi/6}) = \cdots$$

6. Feb 9, 2012

### Alexitron

The square root of a complex number has two solutions.