- #1

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- 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.

##\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.