Integration without using parametrization . I would like to integrate the following line integral without using parametrization . 
I wanted to integrate the following $$\int_C \frac{1}{z-a} dz$$ , where $C$ is a a curve along $|z-a| =r$ . 
Using parametrization i can easily get the answer , but i wanted to integrate without using parametrization of the circle. Appreciate your hints . 
 A: You may assume $a=0$. Consider an integer $N\gg1$, put $\omega:=e^{2\pi i/N}$ (an $N$th root of unity) and choose $z_k:=r\omega^k\in\partial D_r$ $\>(0\leq k\leq N)$. Then using the "physical interpretation" of a line integral one has
$$\eqalign{\int_{\partial D_r}{dz\over z}&\doteq\sum_{k=0}^{N-1}{z_{k+1}-z_k\over z_k}=\sum_{k=0}^{N-1}{r\omega^{k+1}-r\omega^k\over r\omega^k}\cr  &=N(\omega-1)=2\pi i\>{e^{2\pi i/N}-1\over{2\pi i/N}}\to 2\pi i\quad(N\to\infty)\ .\cr}$$
Here is a second approach to the problem:
On the right half plane the principal value $z\mapsto {\rm Log}\>z$ is a primitive of $z\mapsto {1\over z}$. It follows that the integral over the circular arc $\gamma_1$ from $re^{-i\pi/4}$ to $re^{-i\pi/4}$ is given by
$$\int_{\gamma_1}{dz\over z}={\rm Log}\bigl(re^{i\pi/4}\bigr)-{\rm Log}\bigl(re^{i\pi/4}\bigr)=i{\pi\over2}\ .$$
Similarly, on the upper half plane the principal value $z\mapsto {\rm Log}\>{z\over i}$ is a primitive of $z\mapsto {1\over z}$. It follows that the integral over the circular arc $\gamma_2$ from $re^{i\pi/4}$ to $re^{3i\pi/4}$ is given by
$$\int_{\gamma_2}{dz\over z}={\rm Log}{re^{3i\pi/4}\over i}-{\rm Log}{re^{i\pi/4}\over i}=i{\pi\over2}\ .$$
And so on.
