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Prove that $$\frac {1} {2\pi i} \oint \frac {{\rm d}z} {P(z)} $$ over a closed smooth simple curve that contains all of the roots of the polynomial $P(z)$ is either zero if $n \geq 2$ or equals $\frac{1} {a_0} $ if it's degree is $1$.

Proof: $n\geq 2$ $$\frac {1} {P(z)=(z-z_0)(z-z_1)\dots(z-z_n)}$$ So after partial decomposition of the fraction I'll have something like $$\frac {A} {z-z_0} +\frac{B} {z-z_1}+\cdots$$ where $A,B \in \mathbb{C}$ (Is this fact true that $A,B$ wont be polynomials and just coefficients I can see it why in an intuitive way but cannot write a full proof of it for the decomposition trying to get a linear system or something) then use Cauchy's theorem and integrate all of these on different circles with their center at each root and argue that the integral is the same as the sum of those circle integrals. Now since $\frac {1}{z-z_n}$ will be analytic inside the circle when I'm integrating on circles that do not contain $z_n$ some of those integrals will be zero. and now since my curves are of the form $\gamma(t)=z_n+e^{it}$ , $t \in [0,2\pi]$ the rest of the integrals will look like $$\int_0^{2\pi} \frac{Aie^{ti}} {e^{it}}\,{\rm d}t $$ and all of them will be $A_i$.

Is my proof correct? Only part I cannot show is that I won't have any problem with decomposing the fraction. And I'm also using the assumption that after the decomposition I'll have as numerators simple complex numbers not not functions of any kind.

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Let $P(z)=\sum_{k=1}^na_kz^k$ be a polynomial of order $n>1$. Assume that all of the zeros, $z_k$, $k=1,\dots,n$, are distinct. Then, we can write $P(z)=a_n\prod_{k=1}^n(z-z_k)$.

Next, using partial fraction expansion, we can write the reciprocal of $P(z)$ as

$$\begin{align} \frac{1}{P(z)}&=\sum_{k=1}^n\frac{A_k}{z-z_k} \tag 1 \end{align}$$

where $A_k$ are the residues. Now, multiplying $(1)$ by $a_n\prod_{j=1}^n(z-z_j)$ yields

$$\begin{align} \frac{a_n\prod_{j=1}^n(z-z_j)}{P(z)}=a_n\sum_{k=1}^n A_k \frac{\prod_{j=1}^n(z-z_j)}{(z-z_k)} \tag 2 \end{align}$$

The left-hand side of $(2)$ is equal to $1$, whereas the right-hand side is a polynomial of order $n-1$. Therefore, the coefficients of the polynomial on the right-hand side must be zero for all but the zeroth order term. In particular, the coefficient on the leading term is given by $a_n\sum_{k=1}^n A_k$. Since, this must be zero, we find that

$$\sum_{k=1}^n A_k=0$$

Inasmuch as $\oint_C \frac{1}{P(z)}\,dz=2\pi i \sum_{k=1}^n A_k $, then

$$\frac{1}{2\pi i}\oint_C \frac{1}{P(z)}\,dz=0$$

as was to be shown!

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  • $\begingroup$ where does the integral comes in? $\endgroup$ – Manolis Lyviakis Aug 19 '16 at 15:16
  • $\begingroup$ @JackD'Aurizio Thanks! Much appreciative. $\endgroup$ – Mark Viola Aug 19 '16 at 15:17
  • $\begingroup$ @ManolisLyviakis $\frac{1}{2\pi i}\oint_C \frac{1}{P(z)}\,dz=\sum \text{Residues}=\sum_k A_k$. $\endgroup$ – Mark Viola Aug 19 '16 at 15:18
  • $\begingroup$ ohh did not know that. $\endgroup$ – Manolis Lyviakis Aug 19 '16 at 15:19
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    $\begingroup$ @ManolisLyviakis I thought that THIS REFERENCE and THIS ONE would be useful $\endgroup$ – Mark Viola Aug 19 '16 at 16:35
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I think you are over-complicating it a bit, it is not necessary to recover the coefficients of the partial fraction decomposition, it is enough to exploit the residue theorem and the ML lemma. If the degree of $P$ is $n\geq 2$ and all its zeroes lies inside $|z|<R$, $$\forall M\geq R,\qquad \oint_{|z|=R}\frac{dz}{P(z)}=\oint_{|z|=M}\frac{dz}{P(z)} $$ since both integrals equal $2\pi i$ times the sum of the residues of $f(z)=\frac{1}{P(z)}$ at its poles.
But if $|z|$ is large we have $\left|P(z)\right|\geq C |z|^n $, hence $$\left|\oint_{|z|=M}\frac{dz}{P(z)}\right|\leq \frac{2\pi M}{CM^n} $$ that goes to zero as $M\to +\infty$, and the only non-trivial part of the claim is proved.

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  • $\begingroup$ i know it can be done with residue theorem .Everything can be done with residue theorem.haha joking i just wanted to see if my approach is correct and if the decomposition of fraction on C always gives constants on numerators.!!!! $\endgroup$ – Manolis Lyviakis Aug 19 '16 at 13:11
  • $\begingroup$ @ManolisLyviakis: well, the coefficients of the partial fraction decomposition are given by the residues at the poles, so we are following almost the same path. $\endgroup$ – Jack D'Aurizio Aug 19 '16 at 13:12
  • $\begingroup$ I do not know the residue theorem .Could you explain me what do you mean by saying "residues at the poles" , and that we are following the same path? :) $\endgroup$ – Manolis Lyviakis Aug 19 '16 at 13:17
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    $\begingroup$ @ManolisLyviakis: that is quite trivial if you write it down: $$ \lim_{z\to z_1}\left(\frac{A(z-z_1)}{z-z_1}+\frac{B(z-z_1)}{z-z_2}\right)= A$$ $\endgroup$ – Jack D'Aurizio Aug 19 '16 at 13:29
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    $\begingroup$ So according to residues definition these constants are the coefficients of the laurent series of the reciprocal of a polynomial? $\endgroup$ – Manolis Lyviakis Aug 19 '16 at 16:20

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