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Thread: closed curve, trace

  1. #1
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    closed curve, trace

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


    Hint:

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

    Attempt:

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


    Hint:

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

    Attempt:

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

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

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