# Understanding a theorem of Marden's on the moduli of zeros of polynomials

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
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

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