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Math Help - complex logarithm

  1. #1
    Super Member Random Variable's Avatar
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    complex logarithm

    Let  z = x+iy

    I want to show that  \ln(e^{iz}) = 2i \pi \lfloor \frac{1}{2} - \frac{x}{2 \pi} \rfloor + iz

     \ln(e^{iz}) = \ln|e^{iz}| + i \tex{Arg} (e^{iz})

     = \ln (e^{-y}) + i \text{Arg} (e^{iz})

     = -y + i \text{Arg} (e^{iz})


    EDIT:  \text{Arg} (e^{iz}) is some angle in the interval  (-\pi, \pi] . I don't understand how I can express that with use of the floor function.
    Last edited by Random Variable; June 11th 2011 at 04:48 PM.
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  2. #2
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    Quote Originally Posted by Random Variable View Post
    Let  z = x+iy

    I want to show that  \ln(e^{iz}) = 2i \pi \lfloor \frac{1}{2} - \frac{x}{2 \pi} \rfloor + iz

     \ln(e^{iz}) = \ln|e^{iz}| + i \tex{Arg} (e^{iz})

     = \ln (e^{-y}) + i \text{Arg} (e^{iz})

     = -y + i \text{Arg} (e^{iz})


    EDIT:  \text{Arg} (e^{iz}) is some angle in the interval  (-\pi, \pi] . I don't understand how I can express that with use of the floor function.
    Dear Random Variable,

    Whatever the value of x; \lfloor \frac{1}{2} - \frac{x}{2 \pi} \rfloor is an integer. That is,

    \lfloor \frac{1}{2} - \frac{x}{2 \pi} \rfloor=n\mbox{ where }n\in{Z}

    Consider, \exp{\left( 2i \pi \lfloor \frac{1}{2} - \frac{x}{2 \pi} \rfloor+iz\right)}=\exp{(2\pi ni+iz)}=e^{2\pi ni}.e^{iz}=e^{iz}~;~(e^{2\pi ni}=1)

    Hence, \ln(e^{iz}) = 2i \pi \lfloor \frac{1}{2} - \frac{x}{2 \pi} \rfloor + iz
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