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Thread: Limits of complex functions

  1. #1
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    Limits of complex functions

    lim log(sinh(z))
    z->pi i/2
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Richmond View Post
    lim log(sinh(z))
    z->pi i/2
    Hint: $\displaystyle \sinh z = \frac {e^z - e^{-z}}2 = - i \sin iz$
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Richmond View Post
    lim log(sinh(z))
    z->pi i/2
    Note that $\displaystyle \sinh z=\frac{e^z-e^{-z}}{2}$

    So, $\displaystyle \lim_{z\to i\frac{\pi}{2}}\log\left(\sinh z\right)=\log\left(\frac{e^{i\frac{\pi}{2}}-e^{-i\frac{\pi}{2}}}{2}\right)=\log\left(i\right)$

    Since $\displaystyle \log z=\ln\left|z\right|+i\left(\text{Arg}\,z+2k\pi\rig ht) \,z\,;\, -\pi<\text{Arg}\,z\leqslant\pi$, we see that $\displaystyle \log\left(z\right)=\ln \left|i\right|+i\left(\text{Arg}\,i+2k\pi\right)= \ln 1+i\left(\frac{\pi}{2}+2k\pi \right)=\frac{\left(4k+1\right)\pi}{2}i$

    However, I think you're dealing with the Principal Logarithm.

    So, we should say that $\displaystyle \lim_{z\to i\frac{\pi}{2}}\log\left(\sinh z\right)=\boxed{\frac{\pi}{2}i}$

    Does this make sense?
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  4. #4
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    I've got till Log(i)

    I'm not sure what is the Principal Logarithm but I'll try search around the net about it.


    $\displaystyle
    \frac{\left(4k+1\right)\pi}{2}i
    $

    I think this is what I'm looking for. Thanks again for your help. Greatly appreciated. and thanks to Jhevon too. Prefer to work with$\displaystyle - i \sin iz$
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