Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 1)
(→‎Solution 3)
 
(7 intermediate revisions by 5 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
A [[function]] <math>f</math> is defined by <math>f(z) = i\overline{z}</math>, where <math>i=\sqrt{-1}</math> and <math>\overline{z}</math> is the [[complex conjugate]] of <math>z</math>. How many values of <math>z</math> satisfy both <math>|z| = 5</math> and <math>f(z) = z</math>?
+
A function <math>f</math> is defined by <math>f(z) = i\overline{z}</math>, where <math>i=\sqrt{-1}</math> and <math>\overline{z}</math> is the complex conjugate of <math>z</math>. How many values of <math>z</math> satisfy both <math>|z| = 5</math> and <math>f(z) = z</math>?
  
 
<math>\mathrm{(A)}\ 0
 
<math>\mathrm{(A)}\ 0
Line 14: Line 14:
  
 
===Solution 2===
 
===Solution 2===
−
We start the same as the above solution: 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>. Since we are given <math>|z| = 5</math>, this implies that <math>a^2+b^2=25</math>. We recognize the Pythagorean triple <math>3,4,5</math> so we see that <math>(a,b)=(3,4)</math> or <math>(4,3)</math>. So the answer is <math>2 \Rightarrow \mathrm{(C)}</math>.
+
Let <math>z=a+bi</math>, like above. Therefore, <math>z = a+bi = i\overline{z} = i(a-bi) = ai+b</math>. We move some terms around to get <math>bi-b = ai-a</math>. We factor: <math>b(i-1) = a(i-1)</math>. We divide out the common factor to see that <math>b = a</math>. Next we put this into the definition of <math>|z| = a^2 + b^2 = a^2 + a^2 = 2a^2 = 25</math>. Finally, <math>a = \pm\sqrt{\frac{25}{2}}</math>, and <math>a</math> has two solutions.
−
 
 
−
Solution by franzliszt
 
  
 
== See also ==
 
== See also ==

Latest revision as of 12:59, 30 March 2023

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

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 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_12B_Problems/Problem_16&oldid=191570"