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 1

Using the fact $z\bar{z}=|z|^2$ , the equation rewrites itself as

$12z\bar{z}=2(z+2)(\bar{z}+2)+(z^2+1)(\bar{z}^2+1)+31$ $-12z\bar{z}+2z\bar{z}+4(z+\bar{z})+8+z^2\bar{z}^2+(z^2+\bar{z}^2)+32=0$ $\left((z^2+2z\bar{z}+\bar{z}^2)+4(z+\bar{z})+4\right)+\left(z^2\bar{z}^2-12z\bar{z}+36\right)=0$ $(z+\bar{z}+2)^2+(z\bar{z}-6)^2=0.$ As the two quantities in the parentheses are real, $z+\bar{z}=\boxed{\textbf{(A) }-2}$ .

Solution 2

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
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All AMC 12 Problems and Solutions

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2021 AMC 12B Problems/Problem 18

Contents

Problem

Solution 1

Solution 2

See Also