From: Robin Chapman Subject: Re: from Putnam 1989 Date: Thu, 30 Sep 1999 11:10:46 GMT Newsgroups: sci.math Keywords: Polynomial with all roots on unit circle In article <7sv991$3jd$1@draco.tiscalinet.it>, "Giuseppe Macario" wrote: > Prove that if 11z^10 + 10i z^9 + 10i z - 11 = 0, then |z| = 1. It suffices to show that this equation has 10 roots on the unit circle. Setting z = e^{it} makes the LHS 2i e^{5it}(11 sin 5t + 10 cos 4t) so we need to show the equation 11 sin 5t + 10 cos 4t = 0 has 10 real solutions, distinct modulo 2pi. If sin 5t = +- 1 then f(t) = 11 sin 5t + 10 cos 4t has the same sign as sin 5t. Thus f(t) alternates in sign for t = pi/10, 3pi/10, 5pi/10, 7pi/10, 9pi/10, 11pi/10, 13pi/10, 15pi/10, 17pi/10, 19pi/10, 21pi/10 and so there is a solution of f(t) = 0 between each adjacent pair of these. These 10 solutions are distinct modulo 2pi. -- Robin Chapman http://www.maths.ex.ac.uk/~rjc/rjc.html "They did not have proper palms at home in Exeter." Peter Carey, _Oscar and Lucinda_ Sent via Deja.com http://www.deja.com/ Before you buy. ============================================================================== From: Pierre Bornsztein Subject: Re: from Putnam 1989 Date: Thu, 30 Sep 1999 16:03:43 +0200 Newsgroups: sci.math To: Giuseppe Macario Giuseppe Macario wrote: > > Prove that if 11z^10 + 10i z^9 + 10i z - 11 = 0, then |z| = 1. First, remark that z = -10i/11 is not a solution of the equation. let z be a root of 11z^10 + 10i z^9 + 10i z - 11. then : z^9 = (11-10iz)/(11z+10i) let z = a +ib with a,b real numbers. we have |z^9| = f(a,b)/g(a,b) where f(a,b) = sqrt( |11^2 +220b +10^2(a^2 +b^2)| ) and g(a,b) = sqrt( |11^2(a^2 + b^2) + 220b + 10^2 | ) if a^2 + b^2 > 1 : then g(a,b) > f(a,b) thus |z^9| < 1. A contradiction. if a^2 + b^2 < 1 : then g(a,b) < f(a,b) thus |z^9| > 1. A contradiction. then a^2 + b^2 = 1. and we are done. Pierre.