Evaluate $$\int_{|z|=R} \frac{z^{10}-4z^8-6z^3-5}{(z-1)(z-2)(z-5)^9}$$ for all positive $R \neq 1, 2, 5$.

My attempt is to break the solution into four pieces and to apply Cauchy's Integral Theorem.

For $0<R<1$, there are no encircled singularities, so the answer is zero.

For $1<R<2$, only the singularity at $z=1$ is enclosed. The integral is then equal to $$2\pi i\mbox{Res}(1) = 2\pi i *\frac{14}{(-4)^9}$$.

For $2<R<5$, only the singularities at $z=1,2$ are enclosed. The residue at $z=2$ is calculated, added to the above residue for $z=1$, and then multiplied by $2\pi i$ to yield the solution.

Here's my question: For $R>5$, the integral encloses all singularities, and evaluates as $$2\pi i [\mbox{Res}(1)+ \mbox{Res}(2)+\mbox{Res}(5)]$$ Is this correct? If so, how is the residue at $5$ calculated?

This will go a long way to helping me cement complex integration over closed curves. Thanks in advance for any advice!


Yes, what you have so far is correct.

You can compute the residue by differentiating

$$\frac{z^{10} - 4z^8 - 6z^3 - 5}{(z-1)(z-2)}$$

eight times, and plugging $5$ into that, but that becomes rather unwieldy. No problem for a computer algebra system, I think, but by hand, I prefer other methods.

Note that for $R > 5$, the integral is independent of $R$, and computing the limit for $R \to \infty$ is easier: Setting $z = Re^{i\varphi}$, and hence $dz = iz\,d\varphi$, we get

$$\begin{align} \int_{\lvert z\rvert = R} \frac{z^{10} - 4z^8 - 6z^3 - 5}{(z-1)(z-2)(z-5)^9}\,dz &= i\int_0^{2\pi} \frac{z^{11} - 4z^9 - 6z^4 - 5z}{(z-1)(z-2)(z-5)^9}\,d\varphi\\ &= i\int_0^{2\pi} \frac{1 - 4z^{-2} - 6z^{-7} - 5z^{-10}}{(1-z^{-1})(1-2z^{-1})(1-5z^{-1})^9}\,d\varphi. \end{align}$$

In the latter form, it is easy to see that the limit for $R\to\infty$ is $2\pi i$.

  • $\begingroup$ Excellent point at the end. I attempted the derivative formula and wasted time trying to make it work; yours is much more elegant. $\endgroup$ – Darrin Jan 6 '14 at 23:12

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