# Integration in complex plane, circle proof

Show that if $|a| < r <|b|$, then $$\int_\gamma\frac{1}{(z-a)(z-b)}dz=\frac{2\pi i}{a-b}$$ where $\gamma$ denotes a circles about the origin of radius r, with positive orientation. I tried $\gamma (t)=re^{it}$ from 0 to 2pi, but i couldn't seem to work out the answer.

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Use Cauchy's Integral Theorem:

$$\int\limits_\gamma\frac{1}{(z-a)(z-b)}dz=\int\limits_\gamma\frac{\frac{1}{z-b}}{z-a}dz=2\pi i\left.\left(\frac{1}{z-b}\right)\right|_{z=a}=\frac{2\pi i}{a-b}$$

Or using directly residues: the only pole is the simple one is $\,z=a\,$ within $\gamma\,$ , so:

$$Res_{z=a}(f)=\lim_{z\to a}\;(z-a)\frac{1}{(z-a)(z-b)}=\frac{1}{a-b}$$

and again

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

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