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 "2021 AMC 12B Problems/Problem 18"

m (→‎Solution 1)
Line 1: Line 1:
−
==Problem 18==
+
==Problem==
 
Let <math>z</math> be a complex number satisfying <math>12|z|^2=2|z+2|^2+|z^2+1|^2+31.</math> What is the value of <math>z+\frac 6z?</math>
 
Let <math>z</math> be a complex number satisfying <math>12|z|^2=2|z+2|^2+|z^2+1|^2+31.</math> What is the value of <math>z+\frac 6z?</math>
  
Line 7: Line 7:
 
===Solution 1===
 
===Solution 1===
 
The answer being in the form <math>z+\frac 6z</math> means that there are two solutions, some complex number and its complex conjugate. <cmath>a+bi = \frac{6}{a-bi}</cmath> <cmath>a^2+b^2=6</cmath> We should then be able to test out some ordered pairs of <math>(a, b)</math>. After testing it out, we get the ordered pairs of <math>(-1, \sqrt{5})</math> and its conjugate <math>(-1, -\sqrt{5})</math>. Plugging this into answer format gives us <math>\boxed{\textbf{(A) }-2}</math> ~Lopkiloinm
 
The answer being in the form <math>z+\frac 6z</math> means that there are two solutions, some complex number and its complex conjugate. <cmath>a+bi = \frac{6}{a-bi}</cmath> <cmath>a^2+b^2=6</cmath> We should then be able to test out some ordered pairs of <math>(a, b)</math>. After testing it out, we get the ordered pairs of <math>(-1, \sqrt{5})</math> and its conjugate <math>(-1, -\sqrt{5})</math>. Plugging this into answer format gives us <math>\boxed{\textbf{(A) }-2}</math> ~Lopkiloinm
 +
 +
==See Also==
 +
{{AMC12 box|year=2017|ab=B|num-b=17|num-a=19}}
 +
{{MAA Notice}}

Revision as of 19:36, 11 February 2021

Problem

Let $z$ be a complex number satisfying $12|z|^2=2|z+2|^2+|z^2+1|^2+31.$ What is the value of $z+\frac 6z?$

$\textbf{(A) }-2 \qquad \textbf{(B) }-1 \qquad \textbf{(C) }\frac12\qquad \textbf{(D) }1 \qquad \textbf{(E) }4$

Solution

Solution 1

The answer being in the form $z+\frac 6z$ means that there are two solutions, some complex number and its complex conjugate. \[a+bi = \frac{6}{a-bi}\] \[a^2+b^2=6\] We should then be able to test out some ordered pairs of $(a, b)$. After testing it out, we get the ordered pairs of $(-1, \sqrt{5})$ and its conjugate $(-1, -\sqrt{5})$. Plugging this into answer format gives us $\boxed{\textbf{(A) }-2}$ ~Lopkiloinm

See Also

2017 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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=2021_AMC_12B_Problems/Problem_18&oldid=145482"