AIME 2007 I · 第 3 题

AIME 2007 I — Problem 3

专题: Contest Math
难度: L4
来源: AIME

题目详情

Problem

The complex number $z$ is equal to $9+bi$ , where $b$ is a positive real number and $i^{2}=-1$ . Given that the imaginary parts of $z^{2}$ and $z^{3}$ are the same, what is $b$ equal to?

解析

Solution

Squaring, we find that $(9 + bi)^2 = 81 + 18bi - b^2$ . Cubing and ignoring the real parts of the result, we find that $(81 + 18bi - b^2)(9 + bi) = \ldots + (9\cdot 18 + 81)bi - b^3i$ .

Setting these two equal, we get that $18bi = 243bi - b^3i$ , so $b(b^2 - 225) = 0$ and $b = -15, 0, 15$ . Since $b > 0$ , the solution is $\boxed{015}$ .