**Problem**: (360 Problems For Mathematical Contests - 1.1.53) Let

$ P(x) = a_0x^n+a_1x^{n-1}+\cdots+a_n $, $ a_n \neq 0 $,

be a polynomial with complex coefficients such that there is an integer $ m $ with

$ \left| \frac{a_m}{a_n} \right| > nCm $.

Prove that the polynomial $ P $ has at least a zero with absolute value less than $ 1 $.

**Solution**: We will prove the result by contradiction. Assume all the zeros have absolute value (modulus) at least $ 1 $. Let the zeros be $ z_1, z_2, \ldots, z_n $. Our assumption says that $ |z_i| \ge 1 $ for all $ z_i $.

By Vieta's Formulas, we have

$ (-1)^n z_1z_2 \cdots z_n = \frac{a_n}{a_0} $ so $ |z_1z_2 \cdots z_n| = \left| \frac{a_n}{a_0} \right| $.

Also,

$ \displaystyle (-1)^m\sum z_1z_2 \cdots z_m = \frac{a_m}{a_0} $,

where the summation is taken over all sets of $ m $ roots.

Then, by the Triangle Inequality for complex numbers,

$ \displaystyle \left|\frac{a_m}{a_0}\right| = \left| \sum z_1z_2 \cdots z_m \right| \le \sum |z_1z_2 \cdots z_m| $.

But since $ |z_i| \ge 1 $ for all $ z_i $ by our assumption, we know

$ |z_1z_2 \cdots z_n| = |z_1||z_2|\cdots|z_n| \ge |z_1||z_2|\cdots|z_m| = |z_1z_2 \cdots z_m| $

and similarly for all other sets of $ m $ roots. Since there are precisely $ nCm $ sets of $ m $ roots, we have

$ \displaystyle \sum |z_1z_2 \cdots z_m| \le nCm|z_1z_2 \cdots z_n| = nCm\left| \frac{a_n}{a_0} \right| $.

Connecting the inequalities, we find that

$ \left| \frac{a_m}{a_0} \right| \le nCm\left| \frac{a_n}{a_0} \right| $,

which means

$ \left| \frac{a_m}{a_n} \right| \le nCm $,

giving us a contradiction. Hence our original assumption is false and their exists a root $ z_i $ such that $ |z_i| < 1 $ as desired. QED.

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Comment: Once again, Vieta's Formulas are extremely important to know. Also, being able to manipulate the norms of complex numbers and knowing the general properties of them is essential to solving this problem.

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Practice Problem: (360 Problems For Mathematical Contests - 1.1.58) Consider the equation

$ a_0x^n+a_1x^{n-1}+\cdots+a_n = 0 $

with real coefficients $ a_i $. Prove that if the equation has all of its roots real, then $ (n-1)a_1^2 \ge 2na_0a_2 $. Is the reciprocal true?

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