# Complex Analysis question dealing with contour integration

I am working on this problem :

Suppose $z_0$ is any constant complex number interior to any simply closed curve contour C. Show that for all $n>1$, $∫ \frac{dz}{(z-z₀)^n} = 0$.

I know that that if $f$ is holomorphic on the simply connected domain $U$ where $U$ is a subset of $\mathbb{C}$ then $f$ is path-independent. I'm not exactly sure for cases like

$∫_{|z|=2} \frac{dz}{(z-1)^2}$. Here clearly $f(z) = \frac{1}{z-1}$ is analytic everywhere on $\mathbb{C} - \{1\}$ but there's a singularity inside the disc $|z|=2$. From a prior problem, I know that $∫_{C} \frac{dz}{z} = 2\pi i$ and from my understanding this was because $f(z) = \frac{1}{z}$ in this case had a singularity at $z=0$.

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If $f(z)$ is holomorphic inside a simple closed curve $\Gamma$, it is indeed true that $\displaystyle \int_{\Gamma} f(z) dz = 0$.

The above statement doesn't mean that if $f(z)$ is not holomorphic inside a simple closed curve $\Gamma$, then $\displaystyle \int_{\Gamma} f(z) dz \neq 0$.

For instance, lets evaluate an integral similar to the one you have in your question, $I = \displaystyle \int_{|z|=R} \frac{dz}{z^2}$. Note that $f(z)$ is not holomorphic inside $|z| = R$. It has a singularity at $z = 0$.

Setting $z = Re^{i \theta}$, we get that $I = \displaystyle \int_{\theta = 0}^{2 \pi} \frac{R i e^{i \theta} d \theta}{R^2 e^{2i \theta}} = \frac{i}R \displaystyle \int_{\theta = 0}^{2 \pi} e^{-i \theta} d \theta = \frac{i}R \left. \frac{e^{-i \theta}}{-i} \right|_{0}^{2 \pi} = \frac{i}R \frac{1-1}{-i} = 0$.

This is true for $I = \displaystyle \int_{|z|=R} \frac{dz}{z^n}$, since $e^{-2n\pi i} = 1$, where $n \in \{2,3,4,\ldots\}$

However, for $n=1$, note that if $I = \displaystyle \int_{|z|=R} \frac{dz}{z}$ then setting $z = Re^{i \theta}$ gives us $$I = \displaystyle \int_{\theta = 0}^{2 \pi} \frac{R i e^{i \theta} d \theta}{R e^{i \theta}} = i \int_0^{2 \pi} d \theta = 2 \pi i$$

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I did not see it that way. Thanks that was very concrete and my conceptual understanding improved by your clarification. I'm also surprised that you people can type $\LaTeX{}$ very fast. –  Low Scores Mar 14 '12 at 5:31

If $F'(z)=f(z)$ and $\gamma:[a,b]\to\mathbb C$ is a rectifiable curve in the domain of $f$, then $\int_\gamma f(z)dz = F(\gamma(b))-F(\gamma(a))$. In particular, the integral is $0$ if $\gamma$ is a closed curve. When $n>1$, $\frac{1}{(z-z_0)^n}$ has an antiderivative defined everywhere except at $z_0$. (This is not true if $n=1$.)

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