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Homework Help: Imaginary Numbers to Polar form

  1. Jan 6, 2010 #1
    1. The problem statement, all variables and given/known data
    (1+i)i = re

    Find the real values of r and θ.

    3. The attempt at a solution
    Well, after doing a similar(ish) question I decided taking logs would be a good start:

    i loge(1+i) = loger + iθ

    From here, I have no idea where to go. Using a power of i is killing me....
     
  2. jcsd
  3. Jan 6, 2010 #2

    vela

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    Start by writing 1+i in polar form.
     
  4. Jan 6, 2010 #3
    Right. So I should have:

    ([tex]\sqrt{2}[/tex]e([tex]\pi[/tex]/4) i)i

    And from there log?i can't seem to make that get towards a single polar form..
     
    Last edited: Jan 6, 2010
  5. Jan 6, 2010 #4

    vela

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    Sounds like a good plan. Then you can match the real parts on both sides to each other and similarly with the imaginary parts.
     
  6. Jan 6, 2010 #5

    Mark44

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    I don't understand how you got this.

    [tex]1 + i = \sqrt{2}e^{i \pi/4}[/tex]
    [tex]\Rightarrow ln(1 + i) = ln(\sqrt{2}e^{i \pi/4}) = ln\sqrt{2} + ln(e^{i \pi/4})[/tex]

    The last term on the right can be simplified.
     
  7. Jan 6, 2010 #6
    I got to that as the original question was (1+i)i so I had to put it back into the polar form of (1+i). (unless I'm missing something).

    I still can't really see where to go (assuming I've gone the right way).

    i (ln [tex]\sqrt{2}[/tex] ei [tex]\pi [/tex]/4)

    i (ln [tex]\sqrt{2}[/tex] + ln ei [tex]\pi [/tex]/4)
    i (ln [tex]\sqrt{2}[/tex] + i [tex]\pi[/tex] /4 )

    and from there just multiply out to get the imaginary and real parts?
     
  8. Jan 6, 2010 #7

    vela

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    Yup, because on the RHS, the real part is log r and the imaginary part is [tex]\theta[/tex].
     
  9. Jan 6, 2010 #8
    Ah wow, got it! Thanks a load! Great help!
     
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