Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

share|improve this question
    
thanks for editing the format! i didn't know how to format it properly –  matt Sep 30 '12 at 2:40
1  
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
show 2 more comments

2 Answers

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 $p$.

share|improve this answer
    
great! thanks :D –  matt Oct 1 '12 at 3:00
add comment

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.