2021 AMC 12B Problems/Problem 18

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

2021 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.

Page

Toolbox

Search

2021 AMC 12B Problems/Problem 18

Contents

Problem

Solution

Solution 1

See Also