Problem A. 430. (May 2007)
A. 430. Let n2 and let be arbitrary complex numbers with absolute values at most 1 and let
f(x)=(xu_{1})(xu_{2})...(xu_{n}).
Prove that polynomial f'(x) has a root with a nonnegative real part.
(5 pont)
Deadline expired on 15 June 2007.
Solution. If 1 is a multiple root of f then f'(1)=0 and the statement becomes trivial. So we assume that u_{2},...,u_{n}1.
Let the roots of f'(x) be v_{1},v_{2}...,v_{n1}, and consider the polynomial g(x)=f(1x)=a_{1}x+a_{2}x^{2}+...+a_{n}x^{n}.
The nonzero roots of g(x) are 1u_{2},...,1u_{n}. From the Viétaformulas we obtain
The roots of polynomial f'(1x)=g'(x)=a_{1}2a_{2}x...na_{n}x^{n1} are ; from the Viéta formulas again,
Combining the two equations,
For every k, the number u_{k} lies in the unit disc (or on its boundary), and 1u_{k} lies in the circle with center 1 and unit radius (or on its boundary). The operation of taking reciprocals can be considered as the combination of an inversion from pole 0 and mirroring over the real axis. Hence lies in the half plane , i.e. .
Summing up these inequalities,
so at least one lies in the half plane .
Repeating the same geometric steps backwards,
Statistics:
