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Homework Help: Complex Numbers homework

  1. Aug 26, 2007 #1
    1. The problem statement, all variables and given/known data

    Express [tex]cos^{-1}(x+iy)[/tex] in the form [tex]A+iB[/tex]).

    3. The attempt at a solution

    [tex]x=cosa coshb[/tex]
    [tex]y=-sina sinhb[/tex]

    Using these values, I got [tex]x^2+y^2=cos^2a +sinh^2b[/tex], but I dont know where to go from here.

    [tex]2a=cos^{-1}(x^2+y^2 -\sqrt{1-(x+iy)^2}\sqrt{1-(x-iy)^2})[/tex]
    and similarly,

    but after expanding, these expressions are too complex. Is this the final expression though? I dont have the answer, so I have nothing to compare this to.
  2. jcsd
  3. Aug 26, 2007 #2


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    That's not right.
  4. Aug 27, 2007 #3
    Why though? Here's what I did:

    [tex]x+iy=cosa cos(ib) -sina sin(ib)[/tex]
    [tex]x+iy=cosa coshb -i sina sinhb[/tex] (as cos(ib)=cosh b and sin(ib)=i sinhb)
    Equating the real and imaginary parts,
    [tex]x=cos a cosh b[/tex]
    [tex]y=-sina sinhb[/tex]

    Squaring and adding,

    [tex]x^2+y^2=cos^2acosh^2b +sin^2a sinh^2b[/tex]
    As [tex]cosh^2a=1+sinh^2b[/tex],

    [tex]x^2+y^2=cos^2a +sinh^2b(cos^2a+sin^2a)[/tex]
    [tex]x^2+y^2=cos^2a +sinh^2b[/tex].

    What did I do wrong? Where do I go from here?
  5. Aug 30, 2007 #4
    Anyone got any hints/ideas on this one?
  6. Aug 30, 2007 #5


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    I'd do it like this:

    if z = A + iB

    then use the definition of cos...

    [tex]\frac{e^{iz} + e^{-iz}}{2} = x + iy[/tex]

    Then I'd let [tex]h = e^{iz}[/tex]

    So you have

    [tex]\frac{h+1/h}{2} = x + iy[/tex]

    So first solve for h... then get z out of that... then get A and B from that...

    I didn't actually work through this, so I don't know if it will work... just an idea...
  7. Aug 30, 2007 #6


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    What reason do you have for thinking that if x+ iy= cos(a+ bi) then
    x- iy= cos(a-bi)?

    I think I would try to use
    [tex]cos(z)= \frac{e^z+ e^{-z}}{2}[/tex]
    Of course, if z= x+ iy then
    [tex]e^{x+iy}= e^x(cos(y)+ i sin(y))[/tex]
    [tex]e^{-(x-iy)}= e^{-x}x(cos(y)- i sin(y))[/tex]
    so that is
    [tex]cos(x+iy)= \frac{e^x(cos(y)+ i sin(y))+e^{-x}(cos(y)- i sin(y))}{2}[/tex]
    Now separate the real and imaginary parts of that.
  8. Aug 30, 2007 #7


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    But he needs cos^-1(x+iy)...

    Is this derivation to show that x+ iy= cos(a+ bi) implies x- iy= cos(a-bi) ?
  9. Aug 30, 2007 #8
    That's a typo, right?

    There could be two ways to correct it...

    [tex]cosh(z)= \frac{e^z+ e^{-z}}{2}[/tex]


    [tex]cos(z)= \frac{e^{i z}+ e^{-i z}}{2}[/tex]
  10. Aug 30, 2007 #9

    D H

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    It pops right out of the identity,
    [tex]\cos(a+ib) = \cos a \cosh b - i \sin a \sinh b[/tex]
  11. Aug 31, 2007 #10

    Gib Z

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    Letting [tex]\cos z = y = \frac{e^{iz} + e^{-iz}}{2}[/tex], replace all x's with y's and y's with x's, which is what you do to find inverse relations. Then solve for y using logs.
  12. Aug 31, 2007 #11
    Actually, I solved it. Here's what I did:

    [tex]cos(2a)=cos(a+ib+a-ib)=x^2+y^2-\sqrt{(1-(x+iy)^2)(1-(x-iy)^2)}[/tex] (using cos(a+b))
    [tex]cos(2a)=x^2+y^2+(x^2+y^2-1)[/tex] or[tex]cos(2a)=x^2+y^2-(x^2+y^2-1)[/tex]

    This gives you [tex]A=(2n+1)\frac{\pi}{4} or A=\frac{1}{2}cos^{-1}(2x^2+2y^2-1)[/tex]

    [tex] B=\frac{1}{2}cosh^{-1}(2x^2+2y^2-1) or iB=(2n+1)\frac{\pi}{4}[/tex]. But which is it? How do I eliminate one solution set?
    Last edited: Aug 31, 2007
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