# Complex Analysis /Calculus III question

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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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 $\mathbb{C}-\{0\}$.