Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

3 Answers 3

up vote 3 down vote accepted

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 $0$.

share|improve this answer
    
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

Hint: Use Cauchy integral formula.

share|improve this answer

you could let $u=re^{it}-a$ and $du=rie^{it}$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.