losemind

Dear all,

This complex ****ysis problem huanted me for long:

Suppose a polynomial P(z) has all roots in a convex polygon D... how
to prove all the roots of its derivative P'(z) also located in the
same D?

Thanks a lot,

-Lalila

will self

There was a thread on this not long ago. You can find it in Google by
searching for roots derivative polynomial, search only in sci.math.

It was in, or related to that thread that Ignacio mentioned the delightful
fact that if you start with a triangle in the complex plane and form the
polynomial which has its vertices as roots, then the roots of the
derivative of the polynomial are the foci of the Steiner ellipse, the
largest ellipse that can be inscribed in the triangle.

david c. ullrich

Hmm, this one took me a few minutes (actually I came close to
saying it was clearly false...)

Since a convex polygon is the intersection of a family of closed
half-planes it's sufficient to prove the result for D = a closed
half-plane. And now that result follows from this:

Thm. If P is a non-constant polynomial and all the zeroes of P
have strictly positive real part then P'(0) <> 0.

Pf: Say the zeroes are a_1, ... a_n. We can assume that
P is monic: P(z) = (z - a_1)...(z - a_n). Let A be the product
of all the roots.

If you think about it for a second you see that the coefficient
of x in P is plus or minus

A (1/a_1 + ... + 1/a_n).

The fact that a_j has strictly positive real part shows that
the real part of 1/a_j is also strictly positive. Hence the
sum of the 1/a_j is non-zero. So the coefficient of x is
non-zero, hence P'(0) <> 0. QED.


************************

David C. Ullrich

ignacio larrosa cañestro

lalala <losemind@yahoo.com> escribió en el
mensaje|nea4cfeba.0307060827.640dda8c@posting.goog le.com:


It is a consequence of the Luca's theorem:

"If all zeroes of a polynomial P(x) lie in a half plane, then all zeroes of
the derivative P'(x) lie in the same half plane"

We have that is a1, a2, ..., an are the roots of P(x), then

P(z) = c(z - a1)...(z - an)

P'(z)/P(z) = 1/(z - a1) + ... + 1/(z - an).

Suppose that the half Plane H is defined as he part of the plane where
Im((z - a)/b) < 0. If ak is in H and z not, we have then

Im((z - ak)/b) = Im((z - a)/b) - Im((ak - a)/b) > 0

But the imaginary part of reciprocal numbers have opposite sign. Therefore,
under the same assumption,

Im(b/(z - ak)) < 0

If this is true for all k we conclude that

Im(bP'(z)/P(z)) = Sum(Im(b/(z - ak)), k, 1, n) < 0

and consequently P'(z) =/= 0

From: "Complex ****ysis", Lars V. Ahlfors (page 29, 3rd edition)


--
Best Regards,

Ignacio Larrosa Cañestro
A Coruña (España)
ilarrosaQUITARMAYUSCULAS@mundo-r.com