# Square root of negative complex exponential

## Homework Statement

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

## 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!

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.

vela
Staff Emeritus
Homework Helper
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.

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

The square root of a complex number has two solutions.

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