Rational polynomial complex integral: How do work around the singularities?

Question

We are supposed to integrate the polynomial $$\frac{z}{z^3+2z^2-z-2}=\frac{z}{(z+2)(z+1)(z-1)}$$

On a disk of radius $3$ around the origin that contains all roots of the denominator. The title of the exercise strongly implies that we are supposed to use Cauchy's Formula. I've been pushing around linear factors, but can't see how to bring the integrand into the correct form.

How can we bring this polynomial into the form $$\frac{f(z)}{(z-a)^n}$$ , where $f(z)$ is analytic on a disk with radius $3$ around the origin?

Answer 1

How can we bring this polynomial into the form $$\frac{f(z)}{(z−a)^n}$$ , where $f(z)$ is analytic on a disk with radius $3$ around the origin?

You don't. You note that you can integrate around each of the poles separately (due to Cauchy's integral theorem). In other words, do three loops, one around each pole, and use the theorem to show that the sum of the integrals over each of those three is equal to the the integral over a curve that goes around all of the poles at once.

So when integrating around the pole at $z=-2$, you get $$ \frac{z/((z+1)(z-1))}{z+2} $$ which is of the required form. Then you integrate around the pole at $z=-1$, where the function can be given the form $$ \frac{z/((z+2)(z-1))}{z+1} $$ And for the third pole I leave to you to do the rewriting.

---

Edit: Added details about Cauchy's integral theorem: The theorem says that integrating over the two curves in the below two pictures give the same result (the poles of the function are drawn in):

However, in the last one, we can split it into three separate integrals, one over each loop. And since the function only has one pole within each loop, on each loop we can do the required rewriting and use the integral formula to find the value of the integral. Then we just add the three values together afterwards to get the total value of the original integral.

Answer 2

For each of the poles $z_i$ you can find a small disk $B_i := B_{r_i}(z_i)$ with a pole at the center and no other poles inside the disk. Now you have

$$\int_{\vert 3 \vert} \frac{z}{(z+2)(z+1)(z-1)} = \int_{\partial B_{1}} \frac{z}{(z+2)(z+1)(z-1)} dz + \int_{\partial B_{2}} \frac{z}{(z+2)(z+1)(z-1)} dz + \int_{\partial B_{3}} \frac{z}{(z+2)(z+1)(z-1)} dz$$

[to see why this works: lay your smaller circles into the big one and connect them. Since you exclude the poles you can connect them to closed contours and apply Cauchys theorem. It's also quite nice illustrated at the wikipedia example.]

Now you can define $g_1(z) := \frac{z}{(z+2)(z+1)}$ and Cauchys integral formula gives you:

$$\int_{\partial B_1} \frac{z}{(z+2)(z+1)(z-1)} dz = \int_{\partial B_1}\frac{g_1(z)}{z-1} dz = \frac{2\pi i}{1!} g_1(1)$$

For the other integrals define $g_2, g_3$ in a similar way.