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My question is concerning Theorem 3.2 in this paper of Marden's. The gist of the theorem is stated below.

Theorem 3.2.

Every polynomial of the form

$$ f(z) = \sum_{j=0}^{n} (b_j - b_{j-1}) e^{i \theta_j} z^j $$

where

$$ b_{-1} = b_n = 0 < b_0 < b_1 < \cdots < b_{n-1} $$

has all of its zeros in the disk $|z| \leq 1$. Furthermore, every polynomial of the form

$$ g(z) = b_0 + b_1 z + \cdots + b_{n-1} z^{n-1} $$

has all of its zeros in the disk $|z| \leq 1$.

From the way the theorem is worded, it seems like the second part (about the zeros of $g(z)$) follows from the first part. Is this the case?

The theorem is not proved in the paper.

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Maybe this is a job for Schur-Cohn (or Routh-Hurwitz, after performing the customary Möbius transformation from a disk to a half-plane)? –  J. M. Jul 23 '12 at 4:01
    
@J.M. Thanks. Do you know of a reference where I could learn about Schur-Cohn applied to polynomials with complex coefficients? –  Antonio Vargas Jul 23 '12 at 4:13
    
If memory serves, Marden's book ought to have it (I can't check at the moment, since I'm far away from my books). Otherwise, have a look at Henrici's Applied and Computational Complex Analysis; see also this MO thread. –  J. M. Jul 23 '12 at 4:20
1  
Where does $(1-z) g(z)$ have its roots? –  WimC Jul 23 '12 at 5:09
    
@WimC If you copy your comment into the answer box, this question will leave the "unanswered" page. –  user31373 Jul 24 '12 at 0:43

1 Answer 1

up vote 1 down vote accepted

Where does $(1-z)g(z)$ have its roots?

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Thank you, WimC :) –  Antonio Vargas Jul 24 '12 at 22:42

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