Path integral of ${1\over z^{2}}$ around a circle I was wondering how one would go about evaluating
$$\int_C \frac{1}{z^2}dz$$ when $C$ is a circle of positive radius centered at zero.  I suspected the answer should be $2\pi i$, but now I am not so sure.  Could I compare the circle's radius with that of the unit circle, make an annulus, split it into two simply connected regions and subtract the difference to get zero as the answer?
 A: Since you haven't specified any more advanced techniques, I'll assume you're still in the beginning of your complex analysis career.
You parametrize the circle via $z=re^{i\theta}, dz=ire^{i\theta}\,d\theta$ and then you get
$$\int_C{dz\over z^2}=\int_0^{2\pi}{ire^{i\theta}\over r^2e^{2i\theta}}\,d\theta$$
This is then
$${i\over r}\int_0^{2\pi}e^{-i\theta}\,d\theta=0$$
since $e^{-i\theta}$ is periodic with period $2\pi$.
Alternatively, if you remember your vector calculus, you can see that you are integrating $d\left(-{1\over z}\right)$ along a closed path, $\gamma:[a,b]\to \Bbb C$, and since ${1\over z}$ is a smooth path (i.e. a parametrization of your circle), you have that
$$\int_{\gamma}d\left(-{1\over z}\right)={-1\over \gamma(b)}-{-1\over \gamma(a)}=0$$
since $\gamma(b)=\gamma(a)$ and the formula is $\int_\gamma d(f) = f(\gamma(b))-f(\gamma(a))$. (you start and end at the same point).
A: The function $f(z)=\frac{1}{z^2}$ is a meromorphic function with a double pole in $z=0$ with residue zero, hence $\oint_\gamma f(z)\,dz $ for any contour $\gamma$ surrounding the origin is zero by the residue theorem.
