Square root of negative complex exponential

  • Thread starter amalak
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  • #1
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Homework Statement



Solve [itex]\sqrt{-e^{(i2\pi)/3}}[/itex]

Homework Equations





The Attempt at a Solution



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

[itex]\sqrt{-1}[/itex] = i

then,

[itex]e^{(1/2)*(i2\pi)/3}[/itex]

which comes out as: [itex]ie^{i\pi/3}[/itex]

however, the solution is:

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

Answers and Replies

  • #2
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
vela
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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
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So in this case, the correct θ is not [itex] \pi/3 [/itex]? Since, we're trying to solve [itex] \theta/2 [/itex], wouldn't [itex] (2\pi/3)/2 = \pi/3 [/itex] 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:

[itex] -e^{(i2\pi)/3} = e^{-i\pi/3} [/itex], then [itex] \sqrt{e^{-i\pi/3}} = e^{-i\pi/6}[/itex], as desired. This required De Moivre's formula, is there another way to go about this solution?
 
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  • #5
vela
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$$e^{-i\pi/6} = (-i^2)e^{-i\pi/6} = -i(ie^{-i\pi/6}) = \cdots$$
 
  • #6
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The square root of a complex number has two solutions.

The attachment will help you
 

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