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...