Complex polynomial on the unit circle

[hilbert2]

Homework Statement

If $p(z) = a_0 + a_1 z + \dots + a_n z^n$

and $M = \max\left\{|p(z)|:|z|=1\right\}$

then show $|a_k |\leq M$ for all $k=0,1,2,\dots ,n$

Relevant Equations

Complex number $z$ can be written as $z= |z|e^{i\theta}$ with $\theta\in\mathbb{R}$.

So, the values of polynomial $p$ on the complex unit circle can be written as

$\displaystyle p(e^{i\theta}) = a_0 + a_1 e^{i\theta} + a_2 e^{2i\theta} + \dots + a_n e^{ni\theta}$. (*)

If I also write $\displaystyle a_k = |a_k |e^{i\theta_k}$, then the complex phases of the RHS terms of equation (*) are

$\displaystyle\arg \left( a_k e^{ik\theta}\right) = k\theta + \theta_k$.

Now I should somehow choose $\theta$ so that those complex phases are as much on the same half-circle as possible, to make them not cancel out and show that having $|a_k |> M$ for some $k$ and $M = \max\left\{|p(z)|:|z|=1\right\}$ at the same time leads to contradiction.

 

[fresh_42]

Can you use Cauchy's integral formula for the derivatives of $p(z)$?

 

[hilbert2]

Yes, that is probably the way how this is meant to be solved, I just thought it would be a good exercise to solve it some other way. This is not actual homework assigned for me, I just have to refresh my complex analysis skills because of the project I'm working on right now.

 

[fresh_42]

The problem is to separate a specific coefficient, which is best done by differentiation. I first thought about multiplying a factor $|z|^k$, but then we are left with terms $|z|^p$ as well as $|z|^{-q}$ and we don't know whether $|z| \to 0$ or $|z| \to \infty$ is the key. Differentiation kills the lower terms.

If you do not want to use Cauchy, then - as $p(z)$ is smooth - the mean value theorem for $p^{(k)}$ could be an idea.

 

[FactChecker]

If at all possible, you should use those powerful theorems rather than your own ingenuity. They show, and make use of, how constrained analytic functions really are.

 

[hilbert2]

Ok, so the Cauchy integral theorem says

$\displaystyle p(0) = a_0 = \frac{1}{2\pi i}\oint_\gamma\frac{p(z)}{z}dz =\frac{1}{2\pi i} \int_{0}^{2\pi}\frac{p(\gamma (t))}{\gamma(t)}\gamma'(t)dt$

and

$\displaystyle p^{(k)}(0) = k!a_k = \frac{1}{2\pi i}\oint_\gamma\frac{p^{(k)}(z)}{z}dz = \frac{1}{2\pi i}\int_{0}^{2\pi}\frac{p^{(k)}(\gamma(t))}{\gamma(t)}\gamma'(t)dt$,

where the path $\gamma (t)=e^{it}$ for $t\in[0,2\pi]$. This is the counterclockwise path on complex unit circle.

Writing the integral for the case of $a_0$, I have

$\displaystyle a_0 = \frac{1}{2\pi i}\int_{0}^{2\pi}\frac{a_0 + a_1 e^{it} + a_2 e^{2it} + \dots + a_n e^{nit}}{e^{it}}\cdot (ie^{it}) dt \\ \displaystyle = \frac{1}{2\pi}\int_{0}^{2\pi}\left(a_0 + a_1 e^{it} + a_2 e^{2it} + \dots + a_n e^{nit}\right) dt$.

The reason why only $a_0$ is left after the last integration is that the $2\pi$-periodic functions $a_k e^{ikt}$ have a zero integral over one period.

Also, if the integrand $\left(a_0 + a_1 e^{it} + a_2 e^{2it} + \dots + a_n e^{nit}\right)$ is smaller than some upper limit $M>0$ in absolute value for all $t\in [0,2\pi]$, then the result of integration, $a_0$, can't be greater than $2\pi M$ in absolute value either. So this is why an upper limit for $|p(z)|$ on path $\gamma$ also limits the coefficient $a_0$.

Now I just have to do the same for derivatives of $p(z)$... Thanks for the advice.

 

[hilbert2]

The version for the other coefficients $a_k$ is

$\displaystyle a_k = \frac{p^{(k)}(0)}{k!} = \frac{1}{2\pi i}\oint_\gamma \frac{a_0 + a_1 z +\dots +a_n z^n}{z^{k+1}}dz = \\ \displaystyle\frac{1}{2\pi} \int_{0}^{2\pi}\left(a_0 e^{-ikt} + a_1 e^{-i(k-1)t} + \dots + a_n e^{-i(k-n)t}\right)dt \\ \displaystyle = \frac{1}{2\pi}\int_{0}^{2\pi}e^{-ikt}\left(a_0 + a_1 e^{it} + a_2 e^{2it} + \dots + a_n e^{int}\right)dt,$

where in the last integral, the modulus of factor $e^{-ikt}$ is $1$ and the modulus of $(a_0 + a_1 e^{it} + \dots )$ is less than or equal to the upper limit $M$ in the statement of the problem. Therefore, also $a_k \leq M$. $\square$

Edit: and only when $a_k$ is the only coefficient that differs from zero, there's no better upper limit...