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Let $f(z) = \frac{a_0+a_1z+a_2z^2+a_3z^3+.....+a_{n-1}z^{n-1}}{b_0+b_1z+b_2z^2+b_3z^3+.....+b_{n}z^{n}}$ $b_n \ne0$

Assuming zeroes of denominator to be simple, show that sum of residues of $f(z)$ at it's poles is equal to $\frac{a_{n-1}}{b_n}$

I approached the problem by writing denominator into factors say $(z-c_0)(z-c_1)(z-c_2)....(z-c_n)$

and then adding all residues using $\sum_{i=0}^n\lim \limits_{z\to c_n}{(z-c_n)f(z)}$

However I get stuck with different denominators of each term.

I found a solution where it generalises the formula by showing it true for couple of values of $n$. But this doesn't seem satisfactory enough. Is there a simple way to prove the desired result ?

I tried using Cauchy's integral formula but to no avail as well.

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A slightly different approach which can be very useful e.g. when applying the Egorychev method is to use the fact that for a rational function all residues including the residue at infinity sum to zero. (I don't have a reference ready at this time.) The residue at infinity is given by

$$\mathrm{Res}_{z=\infty} f(z) = - \mathrm{Res}_{z=0} \frac{1}{z^2} f(1/z).$$

This means the sum of the residues at the finite poles is given by

$$\mathrm{Res}_{z=0} \frac{1}{z^2} f(1/z)$$

which in the present case yields

$$\mathrm{Res}_{z=0} \frac{1}{z^2} \frac{a_0+a_1/z+\cdots+a_{n-1}/z^{n-1}}{b_0+b_1/z+\cdots+b_n/z^n} \\ = \mathrm{Res}_{z=0} \frac{1}{z^2} \frac{a_0z^n+a_1z^{n-1}+\cdots+a_{n-1}z}{b_0z^n+b_1z^{n-1}+\cdots+b_n} \\ = \mathrm{Res}_{z=0} \frac{1}{z} \frac{a_0z^{n-1}+a_1z^{n-2}+\cdots+a_{n-1}} {b_0z^n+b_1z^{n-1}+\cdots+b_n}.$$

This is

$$[z^0] \frac{a_0z^{n-1}+a_1z^{n-2}+\cdots+a_{n-1}} {b_0z^n+b_1z^{n-1}+\cdots+b_n}$$

because the denominator is not zero at $z=0$ as per the problem definition. With no pole at zero the constant coefficient can be obtained from the ordinary Taylor series and is equal to

$$\left.\frac{a_0z^{n-1}+a_1z^{n-2}+\cdots+a_{n-1}} {b_0z^n+b_1z^{n-1}+\cdots+b_n}\right|_{z=0} = \frac{a_{n-1}}{b_n}.$$

The rule that residues sum to zero may then be used in a variety of creative ways, for example at this MSE link I where it is applied twice, with the residue at infinity vanishing in the second application as per the integral formula. The same technique is used at this MSE link II. The Egorychev method with a non-zero residue at infinity is applied at this MSE limk III (streamlined proof).

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The sum of residues is $$\frac1{2\pi i}\int_C f(z)\,dz$$ where $C$ is a large enough circle with centre $0$. Let the radius $R\to\infty$. $$\frac1{2\pi i}\int_C f(z)\,dz=\frac1{2\pi}\int_0^{2\pi} \frac{\cdots+R^{n-1}a_{n-1}e^{i(n-1)t}}{ \cdots+R^nb_ne^{int}}e^{it}R\,dt$$ etc.

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  • $\begingroup$ How to solve the integral? $\endgroup$ Commented Jun 24, 2017 at 6:25
  • $\begingroup$ You only need its limit as $R\to\infty$ @AbhishekRana $\endgroup$ Commented Jun 24, 2017 at 6:26
  • $\begingroup$ Okay. So that means I can take lim R→∞ inside the integral. Taking R^n common from denominator will make all terms except last 0. Is this allowed? Can you please provide one or two more steps for clarity? $\endgroup$ Commented Jun 24, 2017 at 6:36
  • $\begingroup$ @AbhishekRana I do feel I have given enough hints, but the integrand converges to $a_{n-1}/b_n$ uniformly. $\endgroup$ Commented Jun 24, 2017 at 6:38
  • $\begingroup$ @AnginaSeng This saved me two pages of trying to prove via induction on $n$. Just beautiful $\endgroup$
    – s_a94248
    Commented Aug 12, 2023 at 10:03

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