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I am given 2 pieces of information as below:

1)A polynomial $\displaystyle P(z)=\sum_{n=0}^d a_n z^n$

2) For all $n=0,\dots,d$, $\displaystyle \oint_{|z|=1} \frac{P(z)}{(2z-1)^{n+1}} dz = 0 $

Then I am asked to find what is $P(z)$. Am I supposed to use the residue formula and get the $\mbox{residue}=0$ for all $n=0,\dots,d$?

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Why not using LaTeX? – Did Oct 30 '12 at 8:57
up vote 4 down vote accepted

Hint: As a polynomial of degree at most $d$, $P(z)$ is a linear combination of the polynomials $(2z-1)^n$ for $0\leqslant n\leqslant d$, that is, $P(z)=\sum\limits_{n=0}^da_n(2z-1)^n$ for some complex coefficients $a_n$.

Then, for every $0\leqslant n\leqslant d$, $\displaystyle\oint_{|z|=1}\frac{P(z)}{(2z-1)^{n+1}}\,\mathrm dz=$ $____$ by the residue theorem, hence $a_n=$ $____$ for every $0\leqslant n\leqslant d$.

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These empty little boxes are a nice way to give hints! – Georges Elencwajg Oct 30 '12 at 9:27
Hi. Thanks for the hint. But I still do not get it. How to calculate the first blank? I get very clumsy when working with explicit formula. So I tried to calculate for n=0 first. I get 2(pi)(i)(a0). I'm handicapped at using Latex and it's difficult to express what I do not understand. – newbowl Oct 30 '12 at 10:54
The residue of $P(z)/(2z-1)^{n+1}$ at $z=1/2$ is the sum of the residues of $a_k/(2z-1)^{n+1-k}$. What is the residue of $1/(2z-1)^{n+1-k}$? – Did Oct 30 '12 at 11:01
I am supposed to use the "Limit formula for higher order poles" right? – newbowl Oct 30 '12 at 11:44
But I get residue of 1/(2z−1)^{n+1−k} is zero. – newbowl Oct 30 '12 at 11:55

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