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Thread: 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) $\displaystyle sin(z) = 0$

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

    $\displaystyle e^{iz} = e^{-iz}$

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

    $\displaystyle e^{2iz} = 1$

    $\displaystyle ln(e^{2iz}) = ln(1)$

    $\displaystyle 2iz = 0$

    $\displaystyle z = 0$

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

    2) $\displaystyle e^{iz} = -1$

    $\displaystyle e^{iz} = i^2$

    $\displaystyle ln(e^{iz}) = ln(i^2)$

    $\displaystyle iz = 2ln(i)$

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

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

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