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Homework Help: Complex Roots

  1. Feb 25, 2004 #1

    Zurtex

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    Question:

    Use the formula
    [tex]cos(x + iy) = cos(x)cos(iy) - sin(x)sin(iy)[/tex]
    to find two imaginary numbers whose cosine is 3

    My workings so far:

    [tex]cos(x)cos(iy) - sin(x)sin(iy) = 3 [/tex]
    Therefore:
    [tex]cos(x)cosh(y) - i(sin(x)sinh(y)) = 3[/tex]

    [tex]cos(x)cosh(y) = 3[/tex]
    [tex]sin(x)sinh(y) = 0[/tex]

    [tex]sin(x) = 0[/tex]
    [tex]x = 0[/tex]
    [tex]cos(x) = 1[/tex]
    [tex]cosh(y) = 3[/tex]
    [tex]y = ln(3 + 8^{\frac{1}{2}})[/tex]

    However I can't seem to work out the other root. I know with polynomials it would would be its complex conjugate but I didn't want to assume so for other functions.
     
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  3. Feb 25, 2004 #2

    NateTG

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    I'm not so sure the identities that you're using a valid, but you have:
    [tex]
    \sin(x)\sin(y)=0 \rightarrow \sin(x)=0
    [/tex]
    when [tex]\sin(y)=0[/tex] is also a possibility.
     
  4. Feb 25, 2004 #3

    Zurtex

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    I think you've misread my post.

    [tex]\sin(x)\sinh(y)=0[/tex]
    not
    [tex]\sin(x)\sin(y)=0[/tex]
     
  5. Feb 25, 2004 #4

    NateTG

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    Oh, ok.
    There's also [tex]x=2\pi n[/tex] for varius [tex]n[/tex] right?
     
  6. Feb 25, 2004 #5

    HallsofIvy

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    NateTG: other values for x, first, would not give "imaginary" answers. Since the problem said "imaginary", we could immediately let x= 0 and reduce to cosh(y)= 3. Secondly, other, non-zero values for x require that cosh(y) be less than 1 and that is not possible.


    But cosh is an even function! Yes, it is true that [tex]cosh(ln(3 + 8^{\frac{1}{2}}))= 3[/tex] but it is also true that
    [tex]cosh(-ln(3 + 8^{\frac{1}{2}}))[/tex]= 3.

    The two imaginary numbers whose cosine is 3 are:
    [tex]ln(3 + 8^{\frac{1}{2}})i[/tex] as you got and

    [tex]-ln(3 + 8^{\frac{1}{2}})i[/tex]
     
    Last edited by a moderator: Feb 25, 2004
  7. Feb 25, 2004 #6

    Zurtex

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    I don't know, isn't that something to do with the roots of unity (something we are about to do). I've only just started this stopic and all we have taught so far is that:

    [tex]cos(iz) = cosh(z)[/tex]
    [tex]sin(iz) = i.sinh(z)[/tex]
    [tex]cosh(iz) = cos(z)[/tex]
    [tex]sinh(iz) = i.sin(z)[/tex]
     
  8. Feb 25, 2004 #7

    Zurtex

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    Great thanks :smile: :smile: :smile:
     
  9. Feb 25, 2004 #8

    NateTG

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    Duh... Obviously I should not be posting before donuts and coffee.

    As an aside here's a different way that both answers show up:

    Since [tex]z[/tex] is imaginary, you have:
    [tex]z=0+iy[/tex]
    so
    [tex]3=\cos(z)=\cos(0+iy)=\cos(0)\cos(iy)+\sin(0)\sin(iy)=[/tex]
    [tex]\cos(iy)=\cosh(y)[/tex]
    so
    [tex]3=\cosh(y)[/tex]
    so
    [tex]3=\frac{e^y+e^{-y}}{2}[/tex]
    so let [tex]y'=e^y[/tex]
    then
    [tex]3=\frac{1}{2}(y+\frac{1}{y})[/tex]
    which solves to:
    [tex]y' \in \{3 + \sqrt{8}, 3 - \sqrt {8}\}[/tex]
    so
    [tex]y \in \{\ln(3 + \sqrt{8}), \ln (3 - \sqrt{8})\}[/tex]
    which are the two solutions that you got since
    [tex] -\ln(3 +\sqrt{8})=\ln(\frac{1}{3+\sqrt{8}})=\ln (3-\sqrt{8})[/tex]
     
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