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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!

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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 $2\pi i$.

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  • $\begingroup$ 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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