# Mathematical Food for Thought

Serves a Daily Special and an All-You-Can-Eat Course in Problem Solving. Courtesy of me, Jeffrey Wang.

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Now That’s Efficient. Topic: Algebra/Complex Numbers/Polynomials. Level: Olympiad. March 22nd, 2006

Problem: (360 Problems For Mathematical Contests – 1.1.53) Let

, ,

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

.

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

Solution: We will prove the result by contradiction. Assume all the zeros have absolute value (modulus) at least . Let the zeros be . Our assumption says that for all .

By Vieta’s Formulas, we have

so .

Also,

,

where the summation is taken over all sets of roots.

Then, by the Triangle Inequality for complex numbers,

.

But since for all by our assumption, we know

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

.

Connecting the inequalities, we find that

,

which means

,

giving us a contradiction. Hence our original assumption is false and their exists a root such that 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

with real coefficients . Prove that if the equation has all of its roots real, then . Is the reciprocal true?