Contour integral of quotient of polynomials

Question

I am asked to compute $\displaystyle A = \int_{\partial \mathcal{D}(0,1)} \frac{8z^2-8z+1}{4z^3-8z^2+z-2}dz$.

Let $\displaystyle g(z)= \frac{8z^2-8z+1}{4z^3-8z^2+z-2}$.

I have found $A = 8\pi i$, but when checking on WolframAlpha I get $A=-\pi$. I don't know where my mistake stems from.

Step 1: I find the root of the denominator: $\{-\frac{i}{2}, \frac{i}{2},2\}$. I remark that the moduli of $\frac{i}{2}$ and $-\frac{i}{2}$ are inferior to 1, thus two singularities lie in the contour.

Step 2: I define a function $f_1$ and a contour $C_1$ centered in $z_1=-\frac{i}{2}$. $$\forall z \in C_1, f_1(z)=(z-z_1)g(z)$$

So I have,

$$\forall z \in C_1 \backslash \{z_1\}, g(z)=\frac{f_1(z)}{z-z_1}$$

Step 3: I apply the Cauchy integral theorem:

$$\int_{C_1}g(z)dz=\int_{C_1}\frac{f_1(z)}{z-z_1}dz=2\pi i f_1(z_1)=4\pi i$$

Step 4: I apply the same reasoning and find again $4\pi i$. (Intuitively, by symmetry, it seems that both integrals should be equal. But I may be wrong).

Step 5: I sum the two results obtained at steps 4 and 5, and get $8\pi i$.

I double-checked WolframAlpha and still finds $-\pi$.

Answer 1

Actually, that integral is equal to $2\pi i$. In fact\begin{align}\operatorname{res}_{z=\pm i/2}\left(\frac{8z^2-8z+1}{4z^3-8z^2+z-2}\right)&=\lim_{z\to\pm i/2}\left(z-\left(\pm\frac i2\right)\right)\frac{8z^2-8z+1}{4z^3-8z^2+z-2}\\&=\left.\frac{8z^2-8z+1}{4(z-2)\left(z+\left(\pm\frac i2\right)\right)}\right|_{z=\pm i/2}\\&=\frac12\end{align}and therefore your integral is equal to $2\pi i\left(\frac12+\frac12\right)=2\pi i$.

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Here's another approach, which doesn't use residues. If you decompose$$\frac{8z^2-8z+1}{4z^3-8z^2+z-2}$$into partial fractions, you will get$$\frac{1/2}{z-i/2}+\frac{1/2}{z+i/2}+\frac1{z-2},$$and so, by Cauchy's integral formula, that your integral is equal to $2\pi i\left(\frac12+\frac12\right)=2\pi i$.

Answer 2

Here's another way to approach it. Consider the integral

$$\int_{\partial D(0,R)}\frac{8z^2-8z+1}{4z^3-8z^2+z-2}dz$$

for $R>2$. The integral takes on the same value everywhere, and is only one residue away from the integral we want. Parametrizing and taking the limit we get that

$$\lim_{R\to\infty}\int_{\partial D(0,R)}\frac{8z^2-8z+1}{4z^3-8z^2+z-2}dz = \frac{8i}{4}\cdot 2\pi =4\pi i$$

The residue at $2$ is

$$\lim_{z\to 2}\:(z-2)\cdot\frac{8z^2-8z+1}{(4z^2+1)(z-2)} = 1$$

which means the original integral we wanted was $2\pi i$