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Math Help - Quick question regarding complex exponentials

  1. #1
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    Quick question regarding complex exponentials

    Just wondering, how does e^{-z} - e^{z} = 1 - e^{iz}

    My tutor did this, but I'm not sure how he did it.

    Thanks.
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  2. #2
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    Quote Originally Posted by Glitch View Post
    Just wondering, how does e^{-z} - e^{z} = 1 - e^{iz}
    My tutor did this, but I'm not sure how he did it.
    You need to double cheek your tutor's notes.
    As written that is not true.
    If z=1 the RHS is real but the LHS is not.
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  3. #3
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    It came from this:

    \frac{(e^{-z} - e^{z})}{i(e^{-z} + e^{z})} = 2i

    I'm trying to simplify it.
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  4. #4
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    Quote Originally Posted by Glitch View Post
    It came from this:
    \frac{(e^{-z} - e^{z})}{i(e^{-z} + e^{z})} = 2i
    I'm trying to simplify it.
    What is the goal here?
    Are you trying to solve for z ?
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  5. #5
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    Well the question this involves is:
    Find all solutions z \in C for tan(iz) = 2i

    So I used the exponential form of sin and cos to try and get an expression for tan.
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  6. #6
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    Quote Originally Posted by Glitch View Post
    Well the question this involves is:
    Find all solutions z \in C for tan(iz) = 2i
    This should be a lesson to you. Had you posted the basic question first, you would have had this reply sooner. That reduces to

    e^{-z}-e^{z}=-2e^{-z}-2e^{z}
    1-e^{2z}=-2-2e^{2z}
    3+e^{2z}=0.

    Can you solve that?
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  7. #7
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    Quote Originally Posted by Plato View Post
    This should be a lesson to you. Had you posted the basic question first, you would have had this reply sooner. That reduces to

    e^{-z}-e^{z}=-2e^{-z}-2e^{z}
    1-e^{2z}=-2-2e^{2z}
    3+e^{2z}=0.

    Can you solve that?
    I thought the index laws didn't work for complex exponentials? :/
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