Proving Dieudonne Criterion for Univalent Polynomials

[jem05]

Homework Statement

prove for the dieudonne criterion on univalent polynomials.
polyn. of degree n is univalent in the open disk iff sum(k=1 --> n) a_k sin(kt)/sint z^(k-1)
is not 0 in open unit disk, given 0 < t < pi/2

Homework Equations

hint: Polyn. univ iff [p(ze^it) - p(ze^-it)]/z(e^it - e^-it) =/= 0 in open unit disk

The Attempt at a Solution

all failed attempts. need a hint to begin, thank you all is appreciated.

 

[mighty2000]

Hello there,

The Dieudonne criterion is a useful tool for determining the univalence of polynomials in the open disk. To prove this criterion, we first need to understand the definition of univalence. A polynomial of degree n is said to be univalent in the open disk if it maps the open unit disk onto a convex domain. In simpler terms, this means that the polynomial has only one root in the open unit disk.

Now, let's take a look at the given hint: [p(ze^it) - p(ze^-it)]/z(e^it - e^-it) =/= 0 in open unit disk. This equation can be rewritten as (p(ze^it) - p(ze^-it))/(e^it - e^-it) =/= 0. This is the same as saying that the derivative of the polynomial evaluated at z = 0 is not equal to 0. This is a key observation to keep in mind.

Next, we can use the Taylor series expansion of a polynomial to rewrite the given equation as (p(ze^it) - p(ze^-it))/(e^it - e^-it) = sum(k=1 --> n) a_k sin(kt)/sint z^(k-1). Now, if we assume that the polynomial is univalent in the open disk, we can use the Cauchy integral formula to evaluate the left-hand side of the equation at z = 0. This will give us the derivative of the polynomial at z = 0, which we already know is not equal to 0. This means that the sum on the right-hand side of the equation cannot be equal to 0, since the left-hand side is not equal to 0.

In conclusion, we can see that if the polynomial is univalent in the open disk, then the sum on the right-hand side of the equation is not equal to 0 in the open unit disk. This proves the Dieudonne criterion for univalent polynomials. I hope this helps you in your proof. Good luck!