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Root of a complex number

  • Thread starter kasse
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  • #1
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z^4= 1/2 + i sqrt(3)/2

I start by transforming into polar form:

z^4 = e^(i*Pi/3)

But then I'm blank.
 

Answers and Replies

  • #2
malawi_glenn
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z^4= 1/2 + i sqrt(3)/2

I start by transforming into polar form:

z^4 = e^(i*Pi/3)

But then I'm blank.
Have you tried De Moivre's formula ?
 
  • #3
CompuChip
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I believe the complex root is defined by (that is: it's usually continued from the real function by)
[tex] \sqrt{r e^{i \phi}} = \sqrt{r} e^{i \phi / 2} [/tex]
 
  • #4
malawi_glenn
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I believe the complex root is defined by (that is: it's usually continued from the real function by)
[tex] \sqrt{r e^{i \phi}} = \sqrt{r} e^{i \phi / 2} [/tex]
yes, and from De Moivre's formula we get the general:

[tex] z^{1/n} = r^{1/n}*exp(i \phi / n) [/tex]
 
  • #5
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yes, and from De Moivre's formula we get the general:

[tex] z^{1/n} = r^{1/n}*exp(i \phi / n) [/tex]

So e^((i Pi)/12) is a solution. How about the other three?
 
  • #6
malawi_glenn
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Remember that the same complex number can be obtained by rotating 2*pi and so on.

i.e arg((1/2) + i (rot3 / 2)) = pi / 3 + 2 pi * n, where n is ... -3,-2,-1,0,1,2,3...
 

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