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Math Help - [SOLVED] Complex Problem

  1. #1
    Member ronaldo_07's Avatar
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    [SOLVED] Complex Problem

    Consider the complex function

    f(z) = e^{-iz}.

    (a) Express f(z) in the form u(x, y) + iv(x, y), give u and v.

    (b) Use the Cauchy-Riemann equations and the existence and continuity of the partial derivatives of u and v to show that f '(z) exists for all z.

    (c) Using your results from b), find an expression for f '(z) in terms of z.

    Help with this question please, Thanks.
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  2. #2
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    e^{iz}  = e^{\left( { - y + ix} \right)}  = \left[ {e^{ - y} \cos (x)} \right] + i\left[ {e^{ - y} \sin (x)} \right]
    Now, you finish the other parts.
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  3. #3
    Member ronaldo_07's Avatar
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    Sorry I made a mistake its ie^{-iz}
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    Quote Originally Posted by ronaldo_07 View Post
    Sorry I made a mistake its ie^{-iz}
    So? If you know what e^{-iz} is (what Plato did) then it is easy to find ie^{-iz}.
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  5. #5
    Member ronaldo_07's Avatar
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    ie^{iz}  = e^{\left( { - y + ix} \right)}  = \left[ {e^{ - iy} \cos (x)} \right] + i\left[ {e^{ - iy} \sin (x)} \right]

    is this correct?
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  6. #6
    Member ronaldo_07's Avatar
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    Quote Originally Posted by Plato View Post
    e^{iz}  = e^{\left( { - y + ix} \right)}  = \left[ {e^{ - y} \cos (x)} \right] + i\left[ {e^{ - y} \sin (x)} \right]
    Now, you finish the other parts.

    should it not be y-xi?
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  7. #7
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    Quote Originally Posted by ronaldo_07 View Post
    ie^{iz}  = {\color{red}\large i}e^{\left( { - y + ix} \right)}  = \left[ {{\color{red}\large i}e^{ - iy} \cos (x)} \right] {\color{red}\large -} \left[ {e^{ - iy} \sin (x)} \right]
    You missed a sign and two i's. See the red above.

    Quote Originally Posted by ronaldo_07 View Post
    should it not be y-xi?
    No indeed!
    i(x + yi) = ix - y
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  8. #8
    Member ronaldo_07's Avatar
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    Quote Originally Posted by ronaldo_07 View Post
    -iz} = e^{\left( { - y + ix} \right)} = \left[ {e^{ - iy} \cos (x)} \right] + i\left[ {e^{ - iy} \sin (x)} \right]" alt="ie^{-iz} = e^{\left( { - y + ix} \right)} = \left[ {e^{ - iy} \cos (x)} \right] + i\left[ {e^{ - iy} \sin (x)} \right]" />

    is this correct?
    The - sign was there when i wrote the question origionally.
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