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This is the definition of the fundamental theorem of contour integration that I have:

If $f:D\subseteq\mathbb{C}\rightarrow \mathbb{C}$ is a continuous function on a domain $D \subseteq \mathbb{C}$ and $F:D\subseteq \mathbb{C} \rightarrow \mathbb{C}$ satisfies $F'=f$ on $D$, then for each contour $\gamma$ we have that:

$\int_\gamma f(z) dz =F(z_1)-F(z_0)$

where $\gamma[a,b]\rightarrow D$ with $\gamma(a)=Z_0$ and $\gamma(b)=Z_1$. $F$ is the antiderivative of $f$.

Let $\gamma(t)=Re^{it}, \ 0\le t \le 2\pi, \ R>0$. In my example it said $\int_\gamma \frac{1}{(z-z_0)^2}dz=0$. Im trying to calculate it out myself, but I got stuck.

I get that $f(z)=\frac{1}{(z-z_0)^2}$ has an antiderivative $F(z)=-\frac{1}{(z-z_0)}$.

Thus by the fundamental theorem of contour integration:

$\int_\gamma \frac{1}{(z-z_0)^2}dz =F(z_1)-F(z_0)\\F(\gamma(2\pi))-F(\gamma(0))\\F(Re^{2\pi i})-F(R)\\-\frac{1}{Re^{2\pi i}-z_0} +\frac{1}{R-z_0}\\-\frac{1}{Re^{i}-z_0} +\frac{1}{R-z_0}$

How does $\int_\gamma \frac{1}{(z-z_0)^2}dz=0$?

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From $e^{2\pi i}=1$, doesn't it follow that $F(Re^{2\pi i})-F(R)=F(R)-F(R)=0$? –  Adelaide Dokras May 25 '12 at 0:08
    
Sorry, thanks! My mistake. I mistook $e^{2\pi i} = e^1$ instead of $e^{2\pi i} =1$ –  Derrick May 25 '12 at 0:12
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2 Answers

up vote 2 down vote accepted

The primitive $F(z)$ is one-valued, hence its change over a closed contour is 0. Generally $$\int_C (z-z_0)^m dz = 0$$

where $m\ne-1$ and $$\int_C\frac{dz}{z-z_0}=2\pi i$$ if $z_0$ lies within $C$.

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Thanks for pointing it out, didnt think of it this way! –  Derrick May 25 '12 at 0:31
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$\gamma(2\pi)=Re^{2\pi i}=R=Re^0=\gamma(0)$

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Sorry, thanks! My mistake. I mistook $e^{2\pi i} = e^1$ instead of $e^{2\pi i} =1$ –  Derrick May 25 '12 at 0:13
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