Contour integral of a rational function using Rouche

Question

I would be glad if anyone if could verify that the approach is correct or suggest quicker/better methods for certain parts of the computation. I am trying to solve the integral

$$\int_{|z|=2}{\frac{z^4+1}{z^5+13z-5}}dz$$

• I have shown that the roots of $z^5+13z-5$ are in $\{z\in\mathbb{C} z\leqslant 2\}$ (By Rouche
• I have checked that the roots are simple(By checking it's hcf with the derivative.)
• Obviously the numerator and denominator don't have any common factors so the poles are precisely the zreoes of $z^5+13z-5$
• I get that the residues at $\alpha$ (a root of $z^5+13z-5$) is $$\frac{\alpha^4+1}{5\alpha^4+13}=\frac{\alpha^4+1}{5\alpha^4+13}=\frac{-12\alpha+5}{-52\alpha+25}=\frac{3}{13}-\frac{10}{13(-52\alpha+25)}$$
• I tried to sum $$\frac{10}{13(-52\alpha_k+25)}$$ for $k=1,2,...,5$ and $$\frac{10}{13}\frac{52^4\sum_{j=1,...,5}{\prod_{k\neq j}{(25/52-\alpha_k)}}}{52^5\prod{(25/52-\alpha_k)}}$$
• We now notice that the denominator is $z^5+13z-5$ evaluated at $25/52$ (times a constant) and the numerator is the derivative of $z^5+13z-5$ evaluate at $25/52$. Then I get some horrible arithmetic mess.

I feel the last step seems rather long-winded and I am beginning to wonder if there is a less messier way of doing the summation of the residues.

Answer 1

Since all the poles are inside the contour, it is advantageous to apply the residual theorem to the outside of the contour. To calculate this, apply a coordinate transformation switching inside and outside.

Set $w=2/z$, $dw = -2dz/z^2$ and account for the orientation reversal on the circle to obtain \begin{align} \int_{|z|=2}{\frac{z^4+1}{z^5+13z-5}}dz &=-\int_{|w|=1}{\frac{(2/w)^4+1}{(2/w)^5+13(2/w)-5}}(-2dw/w^2) \\&=\int_{|w|=1}{\frac{2(16+w^4)}{w(32+26w^4-5w^5)}}dw \end{align} which now only has a single pole at $w=0$ so that the whole integral evaluates to $2\pi i$.

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Another way to exploit the absence of poles outside the radius 2 is to set $z=Rw$, $R>2$ to get \begin{align} \int_{|z|=2}\frac{z^4+1}{z^5+13z-5}dz &=\int_{|z|=R}\frac{z^4+1}{z^5+13z-5}dz \\&=\int_{|w|=1}{\frac{R^4w^4+1}{R^5w^5+13Rw-5}}Rdw \\&=\int_{|w|=1} \frac{w^4+\frac1{R^4}}{w^5+\frac{13}{R^4}w-\frac{5}{R^5}}dw\qquad\text{(constant for $R\to\infty$)} \\&=\int_{|w|=1}\frac{dw}w=2\pi i \end{align}

Answer 2

This is not a complete solution, but rather a possible simplification for the calculation.

Notice that $$\int_{|z|=2}{\frac{z^4+1}{z^5+13z-5}}dz = \frac{1}{5}\int_{|z|=2}{\frac{5z^4+13}{z^5+13z-5}}dz - \frac{8}{5}\int_{|z|=2}{\frac{1}{z^5+13z-5}}dz.$$ Let $f(z) = z^5+13z-5$, then the first summand is $$\frac{1}{5}\int_{|z|=2}{\frac{f'(z)}{f(z)}}dz = 2\pi i$$ by the argument principle (and the fact that all the $5$ zeros of $f(z)$ are in the integration region as you proved). Now you are left to find

$$\int_{|z|=2}{\frac{1}{z^5+13z-5}}dz,$$

which you can try to do using the same methods you already used (and without even having to repeat all the computations, actually), but maybe obtaining an easier formula at the end.

If you find a nic solution, please add it to your OP, as I would like to see it.