# On the continuation of a polynomial

This exrcise is from the first section of Marden:

Exercise 12. Let the interior of a piecewise regular curve $C$ contain the origin $\cal O$ and be star-shaped with respect to $\cal O$. If the complex numbers $a_1, a_2, \ldots, a_m$ are given, then $n$ and $b_{m+1}, b_{m+2}, \ldots, b_n$ may be determined so that all of the zeros of

$F(z) = 1 + a_1 z + \cdots + a_m z^m + b_{m+1} z^{m+1} + \cdots + b_n z^n$

lie on $C$.

Hint: Choose the zeros $\zeta_j$ of $G(z) = z^n F(1/z)$ so that $z_j = 1/\zeta_j$ are points of $C$ and so that the Newton-Girard formulas

$s_k + s_{k-1} a_1 + \cdots + k a_k = 0$

are satisfied by the sums $s_r$ of the $r^{\text{th}}$ powers of the $\zeta_j$.

I understand how the requirements of the hint will imply the result, but I do not know how to establish them. Indeed, if we can satisfy those requirements, then the remaining $b_j$ will be determined by the further Newton-Girard formulas.

This is a generalization of the previous problem in the book:

Exercise 11: If $F(z) = 1 + a_1 z + b_2 z^2 + \cdots + b_n z^n$, the quantities $n, b_2, \ldots, b_n$ may be so determined so that all the zeros of F lie on the unit circle.

The solution to this is simpler: we choose $n > -a_1$ and values $\zeta_1, \ldots, \zeta_n$ on the unit circle with centroid $-a_1/n$. Viète's formulas tell us that these are the zeros of a polynomial $z^n + a_1 z^{n-1} + b_2 z^{n-2} + \cdots + b_n = z^n F(1/z)$, so that $1/\zeta_j$ are the zeros of $F(z)$ and lie on the unit circle.

My problem with Exercise 12 lies in ensuring the Newton-Girard formulas are satisfied; it is not as simple as choosing a centroid, and I can't see a way to do it even for the case when $C$ is the unit circle. How do I know that a solution exists here? Can I extend this to the general case, or is a separate argument needed?

Also, how do I center formulas?

Edit: I forgot to mention that Marden gives some citations of this result:

Gavrilov, L., On the continuation of polynomials
Gavrilov, L., On the K-extension of polynomials
Cebotarev, N. G., Über die Fortsetzbarkeit von Polynomen auf geschlassene Kurven
Cebotarev, N. G., On Hurwitz's problem for transcendental functions

Unfortunately I read neither Russian nor German and I can't seem to locate the last one, so these aren't of direct help to me.

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I guess none of the coefficients can be zero then from Cauchy's Theorem. – PEV Jan 14 '11 at 1:42
@Trevor: I don't follow. Which theorem of Cauchy? – Antonio Vargas Jan 14 '11 at 15:05