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Problem A. 430. (May 2007)

A. 430. Let n\ge2 and let u_1=1,u_2,\ldots,u_n be arbitrary complex numbers with absolute values at most 1 and let


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

(5 pont)

Deadline expired on June 15, 2007.

Solution. If 1 is a multiple root of f then f'(1)=0 and the statement becomes trivial. So we assume that u2,...,un\ne1.

Let the roots of f'(x) be v1,v2...,vn-1, and consider the polynomial g(x)=f(1-x)=a1x+a2x2+...+anxn.

The nonzero roots of g(x) are 1-u2,...,1-un. From the Viéta-formulas we obtain

 \sum_{k=2}^m \frac1{1-u_k} = 
\frac{u_2\ldots u_{n-1}+\ldots+u_3\ldots u_n}{u_2\ldots u_n} = 

The roots of polynomial f'(1-x)=-g'(x)=-a1-2a2x-...-nanxn-1 are 1-v_1,\ldots,1-v_{n-1}; from the Viéta formulas again,

 \sum_{\ell=1}^{n-1} \frac1{1-v_\ell} =
\frac{v_1\ldots v_{n-2}+\ldots+v_2\ldots v_{n-1}}{v_1\ldots v_{n-1}} = 

Combining the two equations,

 \sum_{\ell=1}^{n-1} \frac1{1-v_\ell} = 2\sum_{k=2}^m \frac1{1-u_k}.

For every k, the number uk lies in the unit disc (or on its boundary), and 1-uk 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 \frac1{1-u_k} lies in the half plane \mathrm{Re~}z\ge\frac12, i.e. \mathrm{Re}\frac1{1-u_k}\ge\frac12.

Summing up these inequalities,

 \max\limits_{1\le\ell\le n-1} \mathrm{Re}\frac1{1-v_\ell} \ge \frac1{n-1}\sum_{\ell=1}^{n-1} \mathrm{Re}\frac1{1-v_\ell} = \frac2{n-1}\sum_{k=2}^m \mathrm{Re}\frac1{1-u_k} \ge 1,

so at least one \frac1{1-v_\ell} lies in the half plane \mathrm{Re~}z\ge1.

Repeating the same geometric steps backwards,

 \mathrm{Re}\frac1{1-v_\ell}\ge1 \iff \bigg|(1-v_\ell)-\frac12\bigg|\le\frac12 \iff \bigg|v_\ell-\frac12
\bigg|\le\frac12 \Longrightarrow \mathrm{Re}~v_\ell\ge0.


3 students sent a solution.
5 points:Tomon István.
4 points:Nagy 235 János.
Unfair, not evaluated:1 solutions.

Problems in Mathematics of KöMaL, May 2007