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Evaluate the following integral without doing any explicit calculations:

$\int_\gamma \frac{dz}{z^2}$ where $\gamma(t) = \cos(t) + 2i\sin(t)$ for $0 \le t \le 2\pi$.

This exercise comes along with several others that can be solved with Cauchy's Integral Formula. But I don't think it also applies here since both $\frac{1}{z^2}$ and $\frac{1}{z}$ have the same unique singular point.

Hope someone can help.

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    $\begingroup$ Use that $f^{(n)} (a) = \dfrac{n!}{2\pi i} \int_{\gamma} \dfrac{f(z)}{(z-a)^{n+1}} dz$. $\endgroup$ Commented Mar 21, 2014 at 22:23

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$$f(z) = \frac{1}{2 \pi i} \int_{|\xi -z| = r} \frac {f(\xi)}{\xi - z} d\xi$$

So consider $f^\prime(0)$ with $f(\xi) = 1$.

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The Residue Theorem states that for any closed curve $\gamma$ and any meromorphic function $f$: $$ \oint_\gamma f(z)\,dz = 2\pi i\sum_k Res_k(f), $$

and we recall the residues are poles of order one. The function: $$ f(z)=\frac{1}{z^2} $$

is its own Laurent series, and it has residue 0. There are no other poles. Therefore the integral is zero.

If you want to see it more intuitively, try to prove that $1/z^2$ is exactly what you get if you take $-1/w\,\big(1/z-1/(z-w)\big)$ (two opposite residues), and let $w\to 0$...

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The function $f(z):={1\over z^2}$ $(z\in {\mathbb C}\setminus\{0\})$ has the primitive $F(z):=-{1\over z}$. Therefore the integral in question is $=0$, by general principles.

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