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1989 USAMO Problems/Problem 3

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Problem

Let P(z)= z^n + c_1 z^{n-1} + c_2 z^{n-2} + \cdots + c_n be a polynomial in the complex variable z, with real coefficients c_k. Suppose that |P(i)| < 1. Prove that there exist real numbers a and b such that P(a + bi) = 0 and (a^2 + b^2 + 1)^2 < 4 b^2 + 1.

Solution

Let z_1, \dotsc, z_n be the (not necessarily distinct) roots of P, so that P(z) = \prod_{j=1}^n (z- z_j) . Since all the coefficients of P are real, it follows that if w is a root of P, then P( \overline{w}) = \overline{ P(w)} = 0, so \overline{w}, the complex conjugate of \overline{w}, is also a root of P.

Since \lvert i- z_1 \rvert \cdot \lvert i - z_2 \rvert  \dotsm \lvert i - z_n \rvert = \lvert P(i) \rvert < 1, it follows that for some (not necessarily distinct) conjugates z_i and z_j, \lvert z_i-i \rvert \cdot \lvert z_j-i \rvert < 1. Let z_i = a+bi and z_j = a-bi, for real a,b. We note that (a+b+1)^2 - (a+b-1)^2 = 4a + 4b . Thus \begin{align*}(a^2+b^2+1)^2 &= (a^2+b^2-1)^2 + 4a^2 + 4b^2 = \lvert a^2 + b^2 - 1 - 2ai \rvert ^2 + 4b^2 \\&= \lvert ... Since P(a+bi) = P(z_i) = 0, these real numbers a,b satisfy the problem's conditions. \blacksquare

Resources

1989 USAMO (Problems)
Preceded by
Problem 2
1 2 3 4 5 Followed by
Problem 4
All USAMO Problems and Solutions
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