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Math Help - Trigonometric Function (Complex)

  1. #1
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    Trigonometric Function (Complex)

    I have 2 problems that I cannot finish/ not sure.

    Solve for all values of z such that,
    1) sin(z) = 0

    sin(z) = \frac{e^{iz}-e{-iz}}{2i} = 0

    e^{iz} = e^{-iz}

    \frac{e^{iz}}{e^{-iz}} = 1

    e^{2iz} = 1

    ln(e^{2iz}) = ln(1)

    2iz = 0

    z = 0

    However, my problem is z can be \pi or k\pi. What should I do?

    2) e^{iz} = -1

    e^{iz} = i^2

    ln(e^{iz}) = ln(i^2)

    iz = 2ln(i)

    z = \frac{2ln(i)}{i}

    From z = \frac{2ln(i)}{i}, is this a final answer?

    Thank you.
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  2. #2
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    Quote Originally Posted by noppawit View Post
    I have 2 problems that I cannot finish/ not sure.

    Solve for all values of z such that,
    1) sin(z) = 0
    sin(z) = \frac{e^{iz}-e{-iz}}{2i} = 0
    e^{iz} = e^{-iz}
    \frac{e^{iz}}{e^{-iz}} = 1
    e^{2iz} = 1
    ln(e^{2iz}) = ln(1)
    2iz = 0
    z = 0
    In my view, you are misusing the ‘log’ function completely.
    First of all, the notation for the complex form is \log and not \ln.
    The definition: \log(z)=\ln(|z|)+i\arg(z).
    Many of the usual properties do not hold.
    \log(i)= i\left( {\frac{\pi }{2} + 2n\pi } \right) this means that \log \left( {i^2 } \right) \ne 2\log (i).

    But this is true: \log \left( {e^z } \right) = z + 2n\pi i.
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