Value of the path integral $\int_{\partial K(0,1)} \frac {dz} {(z-a)(z-b)}$ along the unit circle $\partial K(0,1)$, where $|a|,|b| < 1$.

Question

I want to determine the value of the path integral along the unit circle $\partial K(0,1)$, where $|a|,|b| < 1$:

$$\int_{\partial K(0,1)} \frac {dz} {(z-a)(z-b)}$$

Assuming $a \neq b$, I compute $\frac {dz} {(z-a)(z-b)} = \frac 1 {a-b}(\frac {-1} {z-b} + \frac 1 {z-a})$.

Thus I get, $\int_{\partial K(0,1)} \frac {dz} {(z-a)(z-b)} = \frac 1 {a-b} (\int_{\partial K(0,1)} \frac {-dz} {z-b} + \int_{\partial K(0,1)} \frac {dz} {z-a})$

However, I've trouble seeing how to proceed. Can someone help me out ?

Answer 1

Then you can use Cauchy's integral formula. See here http://en.wikipedia.org/wiki/Cauchy%27s_integral_formula for a reference. Let $C_1$ be a circle containing $b$ but not $a$, $C_2$ be a circle containing $a$ but not $b$.

$$\frac 1 {a-b} \left(\int_{\partial K(0,1)} \frac {-dz} {z-b} + \int_{\partial K(0,1)} \frac {dz} {z-a}\right)\\=\frac{1}{a-b}\left(\int_{C_1} \frac {-dz} {z-b} + \int_{C_2} \frac {dz} {z-a}\right) \\ =\frac{1}{a-b}(-2\pi i +2\pi i)=0$$

In fact, you don't even have to split up the integral at the beginning. The original integral can be split up into sum of two integrals:

$$\int_{C_1} \frac {dz} {(z-a)(z-b)}+ \int_{C_2} \frac {dz} {(z-a)(z-b)}$$

Since $C_1$ only contains $b$, the first term has $f(z)=\frac{1}{z-b}$ continuous. Similarly $C_2$ only contains $a$, so the second term has $f(z)=\frac{1}{z-a}$ continuous. This gives $$2\pi i (a-b)+2\pi i(b-a)=0$$