Problem 1520 (difficulty: 7/10)

Let \(\displaystyle n\ge2\) and \(\displaystyle u_1=1,u_2,\ldots,u_n\) be complex numbers with absolute value at most \(\displaystyle 1\), and let

\(\displaystyle f(z)=(z-u_1)(z-u_2)\ldots(z-u_n). \)

Show that the polynomial \(\displaystyle f'(z)\) has a root with nonnegative real part.

KöMaL A. 430.

Solution:

If \(\displaystyle 1\) is a multiple root of \(\displaystyle f\), then \(\displaystyle f'(1)=0\) and the statement becomes trivial. So we assume that \(\displaystyle u_2,\ldots,u_n\ne1.\)

Let the roots of \(\displaystyle f'(z)\) be \(\displaystyle v_1,v_2\ldots,v_{n-1}\), and consider the polynomial \(\displaystyle g(z)=f(1-z)=a_1z+a_2z^2+\ldots+a_nz^n\).

The nonzero roots of \(\displaystyle g(z)\) are \(\displaystyle 1-u_2,\ldots,1-u_n\). From the Viéta-formulas we obtain

\(\displaystyle \sum_{k=2}^n \frac1{1-u_k} = \frac{(1-u_2)\ldots (1-u_{n-1})+\ldots+(1-u_3)\ldots (1-u_n)}{(1-u_2)\ldots (1-u_n)} = -\frac{a_2}{a_1}. \)

The roots of the polynomial \(\displaystyle f'(1-z)=-g'(z)=-a_1-2a_2z-\ldots-na_nz^{n-1}\) are \(\displaystyle 1-v_1,\ldots,1-v_{n-1}\); from the Viéta formulas again,

\(\displaystyle \sum_{\ell=1}^{n-1} \frac1{1-v_\ell} = \frac{(1-v_1)\ldots (1-v_{n-2})+\ldots+(1-v_2\ldots v_{n-1})}{(1-v_1)\ldots (1-v_{n-1})} = -\frac{2a_2}{a_1}. \)

Combining the two equations,

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

For every \(\displaystyle k\), the number \(\displaystyle u_k\) lies in the unit disc (or on its boundary), and \(\displaystyle 1-u_k\) lies in the circle with center \(\displaystyle 1\) and unit radius (or on its boundary). The operation of taking reciprocals can be considered as the combination of an inversion from pole \(\displaystyle 0\) and mirroring over the real axis. Hence \(\displaystyle \frac1{1-u_k}\) lies in the half plane \(\displaystyle \mathrm{Re~}z\ge\frac12\), i.e. \(\displaystyle \mathrm{Re}\frac1{1-u_k}\ge\frac12\).

Summing up these inequalities,

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

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

Repeating the same geometric steps backwards,

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

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