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Let $f(z) = A_0 + A_1z + A_2z^2 + \ldots + A_nz^n$ be a complex polynomial of degree $n > 0$.

Show that $\frac{1}{2\pi i} \int\limits_{|z|=R} \! z^{n-1} |f(z)|^2 dz = A_0 \bar{A_n}R^{2n}$.

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thanks for editing the format! i didn't know how to format it properly – matt Sep 30 '12 at 2:40
Write out the conjugate of $f(z)$, multiply by $f(z)$ itself, expand the whole thing, and integrate term-by-term. Most of those integrals are zero. – GEdgar Sep 30 '12 at 3:36
thanks, i tried that, but i end up with all of the integrals being 0. which integrals wouldn't be 0? – matt Sep 30 '12 at 5:16
you end up with some sum of Ai's and conjugate of Ai's as the coefficient for z^q for some integer q. but the integral on a smooth closed curve for z^q is always 0 for q >= 1. – matt Sep 30 '12 at 5:18
if i just look at the Ao*conjugate(An) term. i have it as a coefficient for z^n * z^(n-1) = z^(2n-1). i let z = Re^(it), 0<=t<= 2pi. – matt Sep 30 '12 at 5:47
up vote 1 down vote accepted

Denote by $\bar f$ the polynomial obtained from $f$ by conjugating its coefficients $A_k$. When $|z|^2=z\bar z=R^2$ then $$|f(z)|^2=f(z)\,\overline{f(z)}=f(z)\bar f(\bar z)=f(z)\bar f\Bigl({R^2\over z}\Bigr)\ .$$ Now $$z^{n-1}\bar f\Bigl({R^2\over z}\Bigr)=\bar A_n{R^{2n}\over z} + q(z)\ ,$$ where $q$ is a certain polynomial. It follows that $${1\over2\pi i}\int\nolimits_{\partial D_R} z^{n-1}|f(z)|^2\ dz={1\over2\pi i}\int\nolimits_{\partial D_R} f(z)\Bigl(\bar A_n{R^{2n}\over z} + q(z)\Bigr)\ dz=A_0\bar A_n R^{2n}\ ,$$ because $f(z)=A_0+ z\, p(z)$ for some polynomial $p$.

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great! thanks :D – matt Oct 1 '12 at 3:00

Let $\Gamma = \{z: |z| = R\}$. Recall that $$ \int_{\Gamma} z^k \, dz = \begin{cases} 0 & k \neq -1 \\ 2\pi i & k = -1 \end{cases}$$ Now, when we multiply out $|f(z)|^2$ in terms of $z$ and $\overline{z}$, we are ultimately evaluating an integral of the following form: $$ \frac{1}{2\pi i} \int_{\Gamma} \sum_j B_j z^{p_j} \overline{z}^{k_j} \, dz = \frac{1}{2\pi i} \int_{\Gamma} \sum_j B_j z^{p_j-k_j} R^{2k_j} \, dz$$ for some powers $p_j, k_j$. Then, since we know that $z^k$ integrates to $0$ unless $k = -1$, then we require that $p_j - k_j = -1$. Since the whole integrand is multiplied by $z^{n-1}$ originally, then it must be that $p_j = 0, k_j = n$, so the only term that does not vanish is the one that has the term that was formed from multiplied the $A_0$ term with the $\overline{A}_n\overline{z}^n$ term. Therefore, in summary, $$ \frac{1}{2\pi i} \int_{\Gamma} z^{n-1} |f(z)|^2 \, dz = \frac{1}{2\pi i} \int_{\Gamma} A_0\overline{A}_n R^{2n} z^{-1} \, dz$$ which evaluates to your desired result.

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that works! thanks a bunch :) – matt Oct 1 '12 at 2:54
Sorry, I noticed some typos in my answer, which I will now fix. – Christopher A. Wong Oct 1 '12 at 3:28

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