# Difference between revisions of "2004 AMC 12B Problems/Problem 16"

## Problem

A function $f$ is defined by $f(z) = i\overline{z}$, where $i=\sqrt{-1}$ and $\overline{z}$ is the complex conjugate of $z$. How many values of $z$ satisfy both $|z| = 5$ and $f(z) = z$? $\mathrm{(A)}\ 0 \qquad\mathrm{(B)}\ 1 \qquad\mathrm{(C)}\ 2 \qquad\mathrm{(D)}\ 4 \qquad\mathrm{(E)}\ 8$

## Solutions

### Solution 1

Let $z = a+bi$, so $\overline{z} = a-bi$. By definition, $z = a+bi = f(z) = i(a-bi) = b+ai$, which implies that all solutions to $f(z) = z$ lie on the line $y=x$ on the complex plane. The graph of $|z| = 5$ is a circle centered at the origin, and there are $2 \Rightarrow \mathrm{(C)}$ intersections.

### Solution 2

We start the same as the above solution: Let $z = a+bi$, so $\overline{z} = a-bi$. By definition, $z = a+bi = f(z) = i(a-bi) = b+ai$. Since we are given $|z| = 5$, this implies that $a^2+b^2=25$. We recognize the Pythagorean triple $3,4,5$ so we see that $(a,b)=(3,4)$ or $(4,3)$. So the answer is $2 \Rightarrow \mathrm{(C)}$.

Solution by franzliszt

### Solution 3

Let $z=a+bi$, like above. Therefore, $z = a+bi = i\overline{z} = i(a-bi) = ai+b$. We move some terms around to get $bi-b = ai-a$. We factor: $b(i-1) = a(i-1)$. We divide out the common factor to see that $b = a$. Next we put this into the definition of $|z| = a^2 + b^2 = a^2 + a^2 = 2a^2 = 25$. Finally, $a = \pm\sqrt{\frac{25}{2}}$, and $a$ has two solutions.

## See also

 2004 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 15 Followed byProblem 17 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions

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