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Math Help - Complex Logarithm Function

  1. #1
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    Complex Logarithm Function

    Question:

    Show using a suitable contour that the complex logarithm function is defined by:

    Log(z) = \int_{[1,z]} \frac{1}{\zeta}\hspace{2mm} d\zeta = ln|z| +  i\hspace{1mm}Arg(z)\hspace{5mm} for\hspace{3mm} z \in \mathbb{C}_{\pi} =  \{z : Arg(z) \neq \pi \}

    What I did was i used the contour \gamma (t) = (1-t) + zt \hspace{3mm} 0 \leq t \leq 1 so that:

    \int_{[1,z]} \frac{1}{\zeta} \hspace{2mm}d\zeta = \int_{0}^1  \frac{\gamma'(t)}{\gamma(t)} \hspace{2mm} dt = \int_{0}^1  \frac{z-1}{1-t+zt}\hspace{2mm} dt = ln|z|

    I've reached here and now I'm absolutely stuck. What do i do next?
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  2. #2
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    Quote Originally Posted by garunas View Post
    Question:

    Show using a suitable contour that the complex logarithm function is defined by:

    Log(z) = \int_{[1,z]} \frac{1}{\zeta}\hspace{2mm} d\zeta = ln|z| +  i\hspace{1mm}Arg(z)\hspace{5mm} for\hspace{3mm} z \in \mathbb{C}_{\pi} =  \{z : Arg(z) \neq \pi \}

    What I did was i used the contour \gamma (t) = (1-t) + zt \hspace{3mm} 0 \leq t \leq 1 so that:

    \int_{[1,z]} \frac{1}{\zeta} \hspace{2mm}d\zeta = \int_{0}^1  \frac{\gamma'(t)}{\gamma(t)} \hspace{2mm} dt = \int_{0}^1  \frac{z-1}{1-t+zt}\hspace{2mm} dt = ln|z|

    I've reached here and now I'm absolutely stuck. What do i do next?
    The formula \textstyle\int\frac1\zeta\,d\zeta = \ln|\zeta| is valid for an integral over the positive real line or the negative real line, but it does not work for integrals along complex contours.

    I think that the easiest way to do this question is to use a different contour, namely one that goes along the real axis from 1 to |z| and then around a circular arc from |z| to z. The integral along the straight line segment will then give you \ln|z| without any difficulty. For the circular arc, write z = re^{i\arg z} and parametrise the path by \gamma(\theta) = re^{i\theta} \ (\theta\in[0,\arg z]). Then

    \int_0^{\arg z} \frac{\gamma'(\theta)}{\gamma(\theta)}\,d\theta = \int_0^{\arg z} \frac{ire^{i\theta}}{re^{i\theta}}\,d\theta = i\arg z.
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  3. #3
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    Absolutely amazing!! Cheers!
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