Magyar Information Contest Journal Articles

# 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 10 March 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 $\displaystyle f'(z)\ne0$.

### Statistics:

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

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