Complex Analysis: Show that $\int_{\gamma}\frac{1}{(z-a)(z-b)}dz=\frac{2\pi i}{a-b}$

Question

How can I show that if $|a|<r<|b|$, then

$\int_{\gamma}\frac{1}{(z-a)(z-b)}dz=\frac{2\pi i}{a-b}$,

where $\gamma$ is the circle with center the origin, radius $r$, and positive orientation?

Thanks.

Answer 1

You might want to use Cauchy's integral formula. It says that for any simple, closed curve $\gamma$, and any function $f(z)$ that is holomorphic on the open region $U \subseteq \Bbb C$ bounded by $\gamma$ (and continuous on the closed region $\overline U$ bounded by $\gamma$), then $$ \int_\gamma\frac{f(z)}{(z-\xi)}dz = 2\pi if(\xi) $$ for any $\xi\in U$.

In your case, you want to set $f(z) = \frac{1}{z-b}$, $\xi = a$ and let $\gamma$ be the circle centered at the origin with radius $r$. The formula you're asked to prove should pop out immediately.

Answer 2

Essentially the same approach using the Residue Theorem:

Since $|a|<r<|b|$, the curve $\gamma$ contains one pole, that is $a$.

Let $f(z)=\frac{1}{(z-b)(z-a)}$

So, we have:

$I=\int_{\gamma}\frac{1}{(z-a)(z- b)}dz$

Then, by applying the Residue theorem for $f$ in $γ$, we get:

$I=2\pi{i} Res(f(z),a)=2\pi{i}(lim_{z\rightarrow a} (z-a)f(z))=$ $\frac{2\pi{i}}{a-b}$