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Math Help - Complex Analysis Trigonometric Identity

  1. #1
    Newbie LeraysJeans's Avatar
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    Complex Analysis Trigonometric Identity

    Hello!

    I was asked to show that the following identity holds:

    |cosh(z)|2= sinh2(x) + cos2(y)

    This is what I've got,

    LS= |cosh(z)|2

    = |1/2(ez+e-z)|2

    =1/4(ez+e-z)(ez+e-z)

    =1/4(e2z+2+e-2z)
    RS= sinh2(x) + cos2(y)

    = (1/2)(ex-e-x)(1/2)(ex-e-x)
    + (1/2)(eiy+e-iy)(1/2)(eiy+e-iy)

    = 1/4(e2x-2+e-2x+e2iy+2+e-2iy)

    = 1/4(e2z+e-2z)

    So LS does not equal RS....
    Where is my mistake? Is it in the Absolute value set on the LS? that's the only part that is not intuitive for me...
    Any hints would be killler!!!
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  2. #2
    Senior Member MaxJasper's Avatar
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    Lightbulb Re: Complex Analysis Trigonometric Identity

    |\cosh (z)| = \sqrt{\text{Cos}[y]^2 \text{Cosh}[x]^2+\text{Sin}[y]^2 \text{Sinh}[x]^2}

    then both sides ls & rs expand to:

    \frac{1}{4} e^{-2 z} \left(1+e^{2 z}\right) \left(e^{2 x}+e^{2 i y}\right)
    Last edited by MaxJasper; October 6th 2012 at 05:47 PM.
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  3. #3
    Newbie LeraysJeans's Avatar
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    Re: Complex Analysis Trigonometric Identity

    Thanks MaxJasper,
    I've never seen this expansion of |cosh(z)| before... and not quite sure if I understand it.
    Could you maybe explain the step from left side to right?
    I've only seen |cosh(z)|= (cosh(z)cosh(z))1/2
    I would really appreciate that!

    Edit:
    Are you using trig identities?
    Such as
    cosh(z)=cos(iz)
    cos(iz)= cos(ix)cosh(iy)- (i)sin(ix)sinh(iy)

    Isn't |cosh(z)|= (cosh(z)cosh(z))1/2 a legitimate expansion of |cosh(z)|. So shouldn't it work?
    Last edited by LeraysJeans; October 8th 2012 at 10:26 AM. Reason: New $#!@ has come to light man.
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  4. #4
    Senior Member MaxJasper's Avatar
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    Lightbulb Re: Complex Analysis Trigonometric Identity

    In expansion of |cosh(z)| you must obtain a real value f(x,y) while your method results in another complex number. You say:

    |\cosh (z)|=\frac{1}{4}\left(e^z+e^{-z}\right)^2=....

    while indeed:

    \frac{1}{4}\left(e^z+e^{-z}\right)^2= \frac{1}{4} e^{-2 x} \left(\left(e^{2 x}+1\right) \cos (y)+i \left(e^{2 x}-1\right) \sin (y)\right)^2

    which is another complex number! nice trick

    |\cosh (z)|=real valued f(x,y)= \sqrt{\sinh ^2(x) \sin ^2(y)+\cosh ^2(x) \cos ^2(y)}
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  5. #5
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    Re: Complex Analysis Trigonometric Identity

    LHS: Note that \overline{e^{z}} = e^{\bar{z}}. (Missing that led to your incorrect expansion of \lVert \cosh(z) \rVert^2).

    Proof: e^{\bar{z}} = e^{x - iy} = e^x(\cos(-y) + i \sin(-y)) = e^x(\cos(y) - i \sin(y))

    = \overline{e^x(\cos(y) + i \sin(y))} = \overline{e^{z}}

    ------------------------------------------

    Derivation:

    4 \left\lVert \cosh(z) \right\rVert ^2 = 4 \left\lVert \frac{e^z + e^{-z}}{2} \right\rVert ^2 = \frac{4 \left\lVert e^z + e^{-z}\right\rVert ^2}{4}

    = \left\lVert e^z + e^{-z} \right\rVert ^2 = (e^z + e^{-z})(\overline{e^z + e^{-z}}) = (e^z + e^{-z})(\overline{e^z} + \overline{e^{-z}})

    = (e^z + e^{-z})(e^{\bar{z}} + e^{-\bar{z}}) = e^ze^{\bar{z}} + e^ze^{-\bar{z}} + e^{-z}e^{\bar{z}} + e^{-z}e^{-\bar{z}}

    = e^{z+\bar{z}} + e^{z-\bar{z}} + e^{-z+\bar{z}} + e^{-z-\bar{z}}= e^{2x} + e^{2yi} + e^{-2yi} + e^{-2x}

    = [e^{2x} + e^{-2x} -2] + 2 + [e^{2yi} + e^{-2yi}] = [e^{2x} + e^{-2x} -2] + 2 + 2\cos(2y)

    = [e^{x} - e^{-x}]^2 + 2 + 2(2\cos^2(y) - 1) = 4\left[\frac{e^{x} - e^{-x}}{2}\right]^2 + 2 + (4\cos^2(y) - 2)

    = 4\sinh^2(x) + 4\cos^2(y)
    Last edited by johnsomeone; October 8th 2012 at 02:03 PM.
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  6. #6
    Senior Member MaxJasper's Avatar
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    Lightbulb Re: Complex Analysis Trigonometric Identity

    |\cosh (z)| = \frac{1}{2} [\cosh (2 x)+\cos (2 y)]
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  7. #7
    Newbie LeraysJeans's Avatar
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    Re: Complex Analysis Trigonometric Identity

    AMAZING! Thank you!
    Makes perfect sense!
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