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Math Help - Complex analysis

  1. #1
    Kai
    Kai is offline
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    Complex analysis

    Hi i need help with these 2 questions,

    1) Prove that, if z=x+iy, then cotz = (sin2x-isinh2y)/(cos2y -cos2x)

    i got,....cosz = cosxcoshy - isinxsinhy

    and sinz = sinxcoshy + icosxsinhy

    cotz = cosz/sinz

    = (cosxcoshy - isinxsinhy) / sinxcoshy + icosxsinhy

    Rationalizing and using double angle formula i get

    cotz = (sin2x-isinh2y)/(cos2x-cosh2y)


    2) By considering summation n=1 to N-1 of Z^2n and that Z=e^i@

    show that summation n=1 to N-1 cos2nQ = (sinN@cos(N-1)@)/sin@,...sin@ not equal to zero

    for the second one, i need an idea how to begin it....

    P.S i wrote summation coz i dont know how to put that big E
    Last edited by Kai; September 7th 2008 at 10:14 AM.
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  2. #2
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    Quote Originally Posted by Kai View Post
    Hi i need help with these 2 questions,

    1) Prove that, if z=x+iy, then cotz = (sin2x-isinh2y)/(cos2y -cos2x)

    i got,....cosz = cosxcoshy - isinxsinhy

    and sinz = sinxcoshy + icosxsinhy

    cotz = cosz/sinz

    = (cosxcoshy - isinxsinhy) / sinxcoshy + icosxsinhy

    Rationalizing and using double angle formula i get

    cotz = (sin2x-isinh2y)/(cos2x-cosh2y)


    2) By considering summation n=1 to N-1 of Z^2n and that Z=e^i@

    show that summation n=1 to N-1 cos2nQ = (sinN@cos(N-1)@)/sin@,...sin@ not equal to zero

    for the second one, i need an idea how to begin it....

    P.S i wrote summation coz i dont know how to put that big E
    1) I think your result should be
    \cot(x+iy) = \frac{\sin 2x - i \sinh 2y}{\cosh 2y - \cos 2x}

    2) I can't follow your problem statement. I gather @ means theta, but what is Q (in cos2nQ)?

    Maybe you can restate the problem?
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  3. #3
    Kai
    Kai is offline
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    its cos2n@,..anyway ive been able to solve them , thanks
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