1989 USAMO Problems/Problem 3

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 $w$ , 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 $\begin{align*} (a^2+b^2+1)^2 &= (a^2+b^2-1)^2 + 4a^2 + 4b^2 = \lvert a^2 + b^2 - 1 - 2ai \rvert ^2 + 4b^2 \\ &= \lvert (a-i)^2 - (bi)^2 \rvert^2 + 4b^2 \\ &= \bigl( \lvert a+bi - i \rvert \cdot \lvert a-bi -i \rvert \bigr)^2 + 4b^2 \\ &= \bigl( \lvert z_i - i \rvert \cdot \lvert z_j - i \rvert \bigr)^2 + 4b^2 < 1+4b^2. \end{align*}$ Since $P(a+bi) = P(z_i) = 0$ , these real numbers $a,b$ satisfy the problem's conditions. $\blacksquare$

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1989 USAMO (Problems • Resources)
Preceded by Problem 2	Followed by Problem 4
1 • 2 • 3 • 4 • 5
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1989 USAMO Problems/Problem 3

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