I'm trying to solve this integral:

$$ I = \int\limits_{|z|=r} {z^{n-1} |p(z)|^2 \, dz} \; \big| p(z) = \sum\limits_{k=0}^n a_k z^k,$$

where $a_k$ is constant.

My attempt:


$$ I = \int\limits_0^{2\pi} r^{n-1} e^{in\theta} e^{i\theta} \left|\sum\limits_{k=0}^n a_k r^k e^{ik\theta}\right|^2 r i e^{i\theta} \, d\theta $$

which looks worse but it simplifies a little. Also, as $a_k$ is constant, I assumed that it does not deppend on the coefficent k, which I'm not sure if it's good.

$$ \to i r^n \int\limits_0^{2\pi} e^{in\theta} |a_k|^2 \left|\sum\limits_{k=0}^{n} r^k e^{ik\theta}\right|^2 \, d\theta $$

Considering the geometric sum:

$$ \to i r^n \int\limits_0^{2\pi} e^{in\theta} |a_k|^2 \left| \frac{(1-(re^{i\theta})^k)}{(1-(re^{i\theta}))}\right|^2 \, d\theta $$

And the problem comes when I expand $e^{ix} = \cos(x) + i\sin(x)$, to get the modulus. I get:

$$ \left|\frac{(1-(re^{i\theta})^k)}{(1-(re^{i\theta}))}\right| = \sqrt{\frac{1+r^{2n}-2r^n\cos(n\theta)}{1+r^2 -2r \cos \theta}} $$

which is very similar to Cosine Theorem, i.e.: $a^2=b^2+c^2-2bc\cos\theta$. In this case $b=1$, $c=r$, but I can't deduce how to simplify it more, in order to integrate. I guess it will simplify somewhere somehow.

Any help or hint is highly appreciated.

  • 1
    $\begingroup$ I see the Poisson kernel lurking in this problem, but haven't thought hard enough to say anything insightful. $\endgroup$ – Eric Towers Oct 14 '19 at 1:28
  • 1
    $\begingroup$ Use $|p(z)|^2=p(z) \bar p(z)$ expand all and note that only terms without a (nonzero) power of $e^{i\theta}$ remain as the others cancel by periodicity $\endgroup$ – Conrad Oct 14 '19 at 1:34

Let $w=\frac{z}{r}$ and then $$ \bar{p}(z)=\sum_{k=0}^n\bar{a}_kr^k\bar{w}^k=\sum_{k=0}^n\bar{a}_kr^kw^{-k}. $$ So \begin{eqnarray} I&=&\int\limits_{|z|=r} {z^{n-1} |p(z)|^2 dz}\\ &=&\int_{|w|=1}r^{n}w^{n-1}\sum_{j,k=0}^na_j\bar{a}_kr^{j+k}w^{j-k}dw\\ &=&\int_{|w|=1}r^{n}\sum_{j,k=0}^na_j\bar{a}_kr^{j+k}w^{n-1+j-k}dw\\ &=& \int_{|w|=1}r^{n}(\frac{r^na_0\bar{a}_n}{w}+\cdots)\\ &=&r^{2n}a_0\bar{a}_{n}\cdot2\pi i\\ &=&2\pi ir^{2n}a_0\bar{a}_n. \end{eqnarray} Here $$ \int_{|w|=1}\frac1wdw=2\pi i. $$

  • $\begingroup$ Hi @xpaul, I followed your solution and the only parts I can't yet understand are: when you say $\overline{p(z)} = \sum\limits_{k=0}^{n} \overline{a}_k r^k \overline{w}^k = \sum\limits_{k=0}^{n} \overline{a}_k r^k w^{-k} $why the $-$ sign in $w^{-k}$ ? $\endgroup$ – holahola Oct 14 '19 at 12:46
  • $\begingroup$ Also, when you get to the point where: $I=\int\limits_{|w|=1} r^n a_j \overline{a}_k \sum\limits{k,j=0}^{n} r^{j+k} w^{j-k+n-1} dw$ why is that equal to: $I = \int\limits_{|w|=1} r^n \big( \frac{r^n a_0 \overline{a}_k}{w}+... \big) dw.$ How do you get the sum to be that, and how the rest of the terms banishes? $\endgroup$ – holahola Oct 14 '19 at 12:53
  • $\begingroup$ @StefanQuandt ,this is because $\bar{w}w=1$ and so $\bar{w}=\frac1w$. $\endgroup$ – xpaul Oct 14 '19 at 13:27
  • $\begingroup$ I just realized that the whole sum simplifies getting the integral inside of it, as the only alternative for the integrand to be non-analytic is when the exponent is -1, and that happens when $j=0$ and $k=n$. The rest of the terms will be analytic so their integrals are 0 and banishes. So the integral reduces to Cauchy Integral form. Thanks you very much for all the help; I guess I'm not at the level yet to understand it quickly, but surely the help was very useful. $\endgroup$ – holahola Oct 15 '19 at 2:49

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