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Math Help - closed curve, trace

  1. #1
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    closed curve, trace

    Show that if \gamma is a closed curve in the complex plane, with trace \Gamma, and if z_0 \not \in \Gamma, then \int_{\gamma} \frac{1}{(z-z_0)^n}dz=0, n \geq 2.


    Hint:

    For n \geq 2, (z-z_0)^{-n} has primitive \frac{(z-z_0)^{-n+1}}{1-n} on D.

    Attempt:

    So the hint says that the primitive (the antiderivative) is \frac{(z-z_0)^{-n+1}}{1-n}. However, I do not see how this will help with showing that \int_{\gamma} \frac{1}{(z-z_0)^n}dz=0, n \geq 2. Thanks in advance.
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  2. #2
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    Quote Originally Posted by Erdos32212 View Post
    Show that if \gamma is a closed curve in the complex plane, with trace \Gamma, and if z_0 \not \in \Gamma, then \int_{\gamma} \frac{1}{(z-z_0)^n}dz=0, n \geq 2.


    Hint:

    For n \geq 2, (z-z_0)^{-n} has primitive \frac{(z-z_0)^{-n+1}}{1-n} on D.

    Attempt:

    So the hint says that the primitive (the antiderivative) is \frac{(z-z_0)^{-n+1}}{1-n}. However, I do not see how this will help with showing that \int_{\gamma} \frac{1}{(z-z_0)^n}dz=0, n \geq 2. Thanks in advance.
    This is a special case of a form of the Fundamental Theorem of Calculus applied to the complex plane:

    If f: D \rightarrow \mathbb{C} is continous and such that there exists g: D \rightarrow \mathbb{C} with g'=f on D then for any curve \gamma: [0,1] \rightarrow D we have that \int_{ \gamma } f= g(\gamma (1))-g(\gamma (0)). To prove this (assuming \gamma is C^1 or piecewise so) we just note that \int_{ \gamma } f= \int_{0}^{1} f(\gamma (t)) \gamma ' (t) dt = \int_{0}^{1} g'(\gamma (t)) \gamma ' (t)dt  = \int_{0}^{1} (g \circ \gamma )' (t)dt = g(\gamma (1))-g(\gamma (0))

    PS. I'm naming \gamma and it's trace [MAtH]\Gamma[/tex] as the same.
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