# Complex integral over a circle

In an exercise, I'm supposed to assume that $|a| < r < |b|$ and prove that $\displaystyle\int_{\gamma} \frac{1}{(z - a)(z - b)} dz = \frac{2 \pi i}{a - b}$, where $\gamma$ is a circle of radius $r$ centered at the origin with positive orientation. So I had the idea to express this integral as $\frac{1}{a - b} \displaystyle\int_{\gamma} \frac{1}{z - a} - \frac{1}{z - b} dz$. Then I tried to evaluate each of these separately, but I don't really know what to do. I get $\displaystyle\int_{0}^{2 \pi} \frac{i r e^{it}}{r e^{it} - a} dt$ for the first term. How am I supposed to integrate this? I can't substitute $u = re^{it}$, can I?

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Use the Cauchy integral formula. Also, note that although this function has 2 poles, $a$ and $b$, only $a$ is inside the contour.

If you don't yet know the Cauchy integral formula, I think that $\frac{1}{a-b}\int_\gamma \frac{1}{z-a}-\frac{1}{z-b}\ dz$ is a good start.

$\frac{1}{a-b}\int_\gamma \frac{1}{z-a}\ dz$ is $\frac{2\pi i}{a-b}$ (use the substitution $w=z-a$). The second integral is zero, since the function is holomorphic everywhere. In other words, if we use the definitions to write it out as a real integral plus $i$ times a real integral, we will have conservative vector fields and so the integral around a closed contour is $0$.

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I'm not sure how to use substitution on complex integrals though. This seems to work: $\displaystyle\int_{\gamma} \frac{1}{z - a} dz = \displaystyle\int_{\gamma} \frac{1}{w} dw = \displaystyle\int_{0}^{2 \pi} \frac{i r e^{it}}{r e^{it}} dt = 2 \pi i$, but is that valid? And if it is, I don't see how it can't be used to conclude the same for the other integral. –  Pedro Mar 29 '12 at 2:27
It is valid. All you're doing is translating the complex plane by $a$. Substituting with the other integral doesn't get you anywhere, but you can find a complex anti-derivative and then use the fundamental theorem of calculus to conclude that you get $0$. If you don't see it, it's unfortunately a bit much to explain here. I recommend looking at the first 2 chapters Greene & Krantz (Function Theory of One Complex Variable) for a really good treatment of complex analysis as an outgrowth of multivariable calculus. –  Brett Frankel Mar 29 '12 at 2:32
I just don't see what would be the harm in doing the exact same thing with $b$ in place of $a$. –  Pedro Mar 29 '12 at 9:13
Give it a shot. If you do it right, you should get $0$. –  Brett Frankel Mar 29 '12 at 16:00
When you did your change of variables, you changed the contour. That doesn't matter so long as the old contour is deformable to the new one. But to deform a loop centered at $0$ to a loop centered at $b$, you would need to go through the singularity, which changes the value of the integral. –  Brett Frankel Mar 30 '12 at 0:15
you could let $u=re^{it}-a$ and $du=rie^{it}$