Difference between revisions of "2021 AMC 12B Problems/Problem 18"

Revision as of 00:51, 13 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 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, both quantities must equal $0$ so $z+\frac6z=z+\bar{z}=\boxed{\textbf{(A) }-2}.$

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

Solution 3

Let $x = z + \frac{6}{z}$ . Then $z = \frac{x \pm \sqrt{x^2-24}}{2}$ . From the answer choices， we know that $x$ is real and $x^2<24$ , so $z = \frac{x \pm i\sqrt{24-x^2}}{2}$ . Then we have $|z|^2 = 6$ $|z+2|^2 = (\frac{x}{2} + 2)^2 + \frac{24-x^2}{4} = 2x+10$ $|z^2+1|^2 = |xz -6 +1|^2 = (\frac{x^2}{2}-5)^2 + \frac{x^2(24-x^2)}{4} = x^2 +25$ Plugging the above back to the original equation, we have $12*6 = 2(2x+10) + x^2 + 25 + 31$ $(x+2)^2 = 0$ So $x = \boxed{\textbf{(A) }-2}$ .

~Sequoia

@@ Line 21: / Line 21: @@
 <cmath>  |z+2|^2 = (\frac{x}{2} + 2)^2 + \frac{24-x^2}{4} = 2x+10</cmath>
 <cmath>  |z^2+1|^2 = |xz -6 +1|^2 = (\frac{x^2}{2}-5)^2 + \frac{x^2(24-x^2)}{4} = x^2 +25</cmath>
-Plug the above back to the original equation, we have
+Plugging the above back to the original equation, we have
 <cmath> 12*6 = 2(2x+10) + x^2 + 25 + 31</cmath>
 <cmath> (x+2)^2 = 0</cmath>
-So <math>x = -2</math>  <math>\boxed{\textbf{(A) }-2}</math>.
+So <math>x = \boxed{\textbf{(A) }-2}</math>.
 ~Sequoia

2021 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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