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I want to integrate $\int_\gamma \! \frac{dz}{z^2}\,$ where $\gamma(t) = e^{it}$ from $0 \leq t \leq \pi$ which is the top half of the unit circle. I keep on getting zero as an answer. Does this mean that the function $f(z) = \frac{1}{z^2}$ is analytic everywhere on the curve $\gamma$?

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up vote 2 down vote accepted

You might want to check your calculation. Integrating over $\gamma=e^{it}$, letting $z=e^{it}$ and $dz = ie^{it}$, we get that we're really integrating $$i\int_0^{\pi}e^{-it}dt$$ which comes out to 2.

Now, to your other question-- checking whether a complex line integral over a path is not sufficient for saying a function is analytic over a curve. In fact, $f(z)=1/z^2$ is analytic over all of $\mathbb{C}-\{0\}$.

Morera's theorem states that if integrating your function over any simple closed curve in your domain gives you zero, then you have an analytic function in that domain. As the path you have chosen is not closed, your integral tells you very little in fact.

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Oh yes you are right it is 2. I checked my calculations. Thanks. –  CalculusStudent Mar 9 '12 at 4:02

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