Mathematical and Physical Journal
for High Schools
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Problem A. 609. (February 2014)

A. 609. Let \(\displaystyle a_1,a_2,\dots,a_n\) and \(\displaystyle b_1,b_2,\dots,b_n\) be complex numbers satisfying \(\displaystyle \mathop{\rm Im} a_j\ge 1\) and \(\displaystyle \mathop{\rm Im} b_j\le -1\) (\(\displaystyle j=1,2,\dots,n\)), and let \(\displaystyle f(z) = \frac{(z-a_1)(z-a_2)\dots (z-a_n)}{(z-b_1)(z-b_2)\dots (z-b_n)}\). Prove that the function \(\displaystyle f'(z)\) has no root in the set \(\displaystyle |\mathop{\rm Im}z|<1\).

(5 pont)

Deadline expired on March 10, 2014.

Sorry, the solution is available only in Hungarian. Google translation

Megoldásvázlat. Ha \(\displaystyle |\mathop{\rm Im}z|<1\), akkor

\(\displaystyle \mathop{\rm Im} \frac{f'(z)}{f(z)} = \mathop{\rm Im} \left( \sum_{k=1}^n \frac1{z-a_k} - \sum_{k=1}^n \frac1{z-b_k} \right) >0, \)

így \(\displaystyle f'(z)\ne0\).


6 students sent a solution.
5 points:Fehér Zsombor, Janzer Barnabás, Kabos Eszter, Maga Balázs, Szőke Tamás, Williams Kada.

Problems in Mathematics of KöMaL, February 2014