# Complex polynomial on the unit circle

• hilbert2
In summary, the values of a polynomial on the complex unit circle can be written in terms of its coefficients. By choosing a specific value for the argument of the coefficients, it is possible to ensure that the complex phases of the terms on the right-hand side of the equation are not canceled out and lead to a contradiction. This can be done using Cauchy's integral formula for the derivatives of the polynomial. Additionally, the mean value theorem for derivatives can also be used to solve the problem. By applying the Cauchy integral theorem, it can be shown that an upper limit for the modulus of the polynomial on the unit circle also limits the coefficients of the polynomial. This is because the integrand of the Cauchy integral is a periodic
hilbert2
Gold Member
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.

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.

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.

hilbert2
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.

Last edited:
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...

Last edited:

## What is a complex polynomial?

A complex polynomial is an algebraic expression with one or more complex variables and coefficients. It is made up of terms that consist of a constant multiplied by a variable raised to a non-negative integer power.

## What is the unit circle?

The unit circle is a circle with a radius of 1 unit, centered at the origin (0,0) on a Cartesian coordinate system. It is commonly used in mathematics to represent complex numbers and their geometric properties.

## How are complex polynomials related to the unit circle?

A complex polynomial on the unit circle is a polynomial function that has roots (or zeros) located on the unit circle. This means that when plotted on the complex plane, the points where the polynomial crosses the unit circle will be the solutions to the polynomial equation.

## What is the significance of complex polynomials on the unit circle?

Complex polynomials on the unit circle have important applications in fields such as signal processing, control theory, and electrical engineering. They also have connections to other areas of mathematics, such as complex analysis and number theory.

## How can complex polynomials on the unit circle be graphed?

Complex polynomials on the unit circle can be graphed by plotting the roots of the polynomial on the complex plane. The points where the polynomial crosses the unit circle will represent the solutions to the polynomial equation. Alternatively, the polynomial can be graphed using parametric equations, where the real and imaginary parts of the polynomial are plotted separately on the x- and y-axes.

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