1999 AIME Problems/Problem 9

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Problem

A function $f$ is defined on the complex numbers by $f(z)=(a+bi)z,$ where $a_{}$ and $b_{}$ are positive numbers. This function has the property that the image of each point in the complex plane is equidistant from that point and the origin. Given that $|a+bi|=8$ and that $b^2=m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers, find $m+n.$

Solution

Solution 1

Suppose we pick an arbitrary point on the complex plane, say $(1,1)$ . According to the definition of $f(z) = f(1+i) = (a+bi)(1+i) = (a-b) + (a+b)i$ , this image must be equidistant to $(1,1)$ and $(0,0)$ . Thus the image must lie on the line with slope $-1$ and which passes through $\left(\frac 12, \frac12\right)$ , so its graph is $x + y = 1$ . Substituting $x = (a-b)$ and $y = (a+b)$ , we get $2a = 1 \Rightarrow a = \frac 12$ .

By the Pythagorean Theorem, we have $\left(\frac{1}{2}\right)^2 + b^2 = 8^2 \Longrightarrow b^2 = \frac{255}{4}$ , and the answer is $\boxed{259}$ .

Solution 2

We are given that $(a + bi)z$ is equidistant from the origin and $z.$ This translates to $\begin{eqnarray*} |(a + bi)z - z| & = & |(a + bi)z| \\|z(a - 1) + bzi| & = & |az + bzi| \\|z||(a - 1) + bi| & = & |z||a + bi| \\(a - 1)^2 + b^2 & = & a^2 + b^2 \\& \Rightarrow & a = \frac 12 \end{eqnarray*}$ Since $|a + bi| = 8,$ $a^2 + b^2 = 64.$ But $a = \frac 12,$ thus $b^2 = \frac {255}4.$ So the answer is $259$ .

1999 AIME (Problems • Resources)
Preceded by Problem 8	Followed by Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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