$\int_{|z| = 2 }\frac{4z^7-1}{z^8-2z+1} dz= ?$

Question

Problem: Compute $\int_{|z|=2} \frac{4z^7 - 1}{z^8 - 2z + 1}dz$.

Attempt: I notice that if $w = z^8 - 2z + 1$ then $dw = 2 (4z^8 - 1) dz$ so $$I=\int_{|z|=2} \frac{4z^7 - 1}{z^8 - 2z + 1}dz = \frac{1}{2}\int_C \frac{1}{w}dw$$ over the closed curve $C = \{ (2e^{i \theta})^8 - 2 (2e^{i \theta}) + 1 \in \mathbb{C} : 0 \leq \theta \leq 2 \pi \}$.

At this point, I know that $\int_C \frac{1}{w} dw = 2\pi i \cdot (\text{winding number of }C\text{ around w = 0})$ so $I = \frac{1}{2} 2\pi i \cdot 8 = 8 \pi i$. Is this not correct?

When I try to verify my answer using Wolfram Alpha, I can get a complex number with nonzero real part: Wolfram link

Maybe Wolfram Alpha is just approximating this integral up to some error bound so their approximation terminates without realizing the real part is actually $0$?

Answer 1

The residue at infinity serves as a sort of "dual" to the residue at zero. Namely,

$$\mathrm{Res}(f(z), \infty) = - \mathrm{Res}\left(\frac{1}{z^2}f\left(\frac{1}{z} \right), 0 \right). $$

Recall that the residue theorem looks at the contributions of singularities inside the contour. But $\infty$ is outside of our contour. How do we fix this? Easy - reverse the orientation. That's where the minus sign comes in.

Let $f(z) = \frac{4z^7 - 1}{z^8 - 2z + 1}.$ Then

\begin{align*} f\left(\frac{1}{z} \right) &= \frac{\frac{4}{z^7} - 1}{\frac{1}{z^8} - \frac{2}{z} + 1} \\ &= \frac{4z - z^8}{1 - 2z^7 + z^8} \end{align*}

Computing the needed value is easy. Thanks to some nice cancellation,

$$ \lim_{z \to 0} z\cdot \frac{1}{z^2} f\left(\frac{1}{z} \right) = 4 $$

and $\mathrm{Res}(f(z), \infty) = -4.$

Let $\gamma$ be the circle of radius $2$ centered at the origin (our contour). Now since $\infty$ is outside $\gamma$, we instead integrate over $-\gamma$ by reversing orientation so that $\infty$ is inside $-\gamma.$ So,

\begin{align*} \oint\limits_{\gamma} f(z) \ dz &= -\oint\limits_{-\gamma} f(z) \ dz \\ &= -2\pi i \mathrm{Res}(f(z), \infty) \\ &= 8\pi i \end{align*}

as needed!

Answer 2

An expedient route is to make use of Rouche and the Argument Principle.

With $\gamma:[0,1]\longrightarrow \mathbb C$ given by $\gamma(t)=2\cdot\exp\big(2\pi i\cdot t\big)$ and
$p(z):=z^8 -2z +1$, notice for any $z$ with modulus $2$
$\big\vert z \big\vert^8 \gt 5 \geq \big\vert -2\cdot z +1\big\vert $ so $p(z)$ and $z^8$ have the same number of roots in $B\big(0,2\big)$ per Rouche.

Then the Argument Principle says
$8 =n\big(p\circ\gamma,0\big) = \frac{1}{2\pi i}\int_\gamma\frac{p'(z)}{p(z)} dz$
$\implies 8 \pi i=\int_{\gamma} \frac{4z^7 - 1}{z^8 - 2z + 1}dz$