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

Revision as of 09:47, 8 July 2020

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

2004 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

@@ Line 8: / Line 8: @@
 \qquad\mathrm{(E)}\ 8</math>
-== Solution 1==
+== Solutions==
+===Solution 1===
 Let <math>z = a+bi</math>, so <math>\overline{z} = a-bi</math>. By definition, <math>z = a+bi = f(z) = i(a-bi) = b+ai</math>, which implies that all solutions to <math>f(z) = z</math> lie on the line <math>y=x</math> on the complex plane. The graph of <math>|z| = 5</math> is a [[circle]] centered at the origin, and there are <math>2 \Rightarrow \mathrm{(C)}</math> intersections.

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