**A. 430.** Let *n*2 and let be arbitrary complex numbers with absolute values at most 1 and let

*f*(*x*)=(*x*-*u*_{1})(*x*-*u*_{2})...(*x*-*u*_{n}).

Prove that polynomial *f*'(*x*) has a root with a non-negative real part.

(5 points)

**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*_{n-1}, and consider the polynomial *g*(*x*)=*f*(1-*x*)=*a*_{1}*x*+*a*_{2}*x*^{2}+...+*a*_{n}*x*^{n}.

The nonzero roots of *g*(*x*) are 1-*u*_{2},...,1-*u*_{n}. From the Viéta-formulas we obtain

The roots of polynomial *f*'(1-*x*)=-*g*'(*x*)=-*a*_{1}-2*a*_{2}*x*-...-*na*_{n}*x*^{n-1} 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 1-*u*_{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,