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 $\begin{align*} 12z\bar{z}&=2(z+2)(\bar{z}+2)+(z^2+1)(\bar{z}^2+1)+31 \\ -12z\bar{z}+2z\bar{z}+4(z+\bar{z})+8+z^2\bar{z}^2+(z^2+\bar{z}^2)+32&=0 \\ \left((z^2+2z\bar{z}+\bar{z}^2)+4(z+\bar{z})+4\right)+\left(z^2\bar{z}^2-12z\bar{z}+36\right)&=0 \\ (z+\bar{z}+2)^2+(z\bar{z}-6)^2&=0. \end{align*}$ 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 $|z|^2 = 6$ $|z+2|^2 = (\frac{x}{2} + 2)^2 + \frac{24-x^2}{4} = 2x+10$ $|z^2+1|^2 = |xz -6 +1|^2 = \left(\frac{x^2}{2}-5\right)^2 + \frac{x^2(24-x^2)}{4} = x^2 +25$ 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

Solution 4 (funny observations)

There are actually several ways to see that $|z|^2 = 6.$ I present two troll ways of seeing it, and a legitimate way of checking.

Rewrite using $w \overline{w} = |w|^2$

$12z \overline{z} + 2(z+2)(\overline{z} + 2) + (z^2+1)(\overline{z}^2+1)+31$ $12 z \overline{z} = 2z \overline{z} + 4z + 4 \overline{z} + 8 + z^2 \overline{z}^2+z^2+\overline{z}^2 + 1 + 31.$ $12 z \overline{z} = 4(z + \overline{z}) + (z \overline{z})^2 + (z + \overline{z})^2 + 40.$

Symmetric in $z$ and $\overline{z},$ so if $w$ is a sol, then so is $\overline{w}$

TROLL OBSERVATION #1: ALL THE ANSWERS ARE REAL. THUS, $z + \frac{6}{z} \in \mathbb{R},$ which means they must be conjugates and so $|z|^2 = 6.$

TROLL OBSERVATION #2: Note that $z+\frac{6}{z} = \overline{z} + \frac{6}{\overline{z}}$ because either solution must give the same answer! which means that $|z|^2 = 6.$

Alternatively, you can check: Let $a = w + \overline{w} \in \mathbb{R},$ and $r = |w|^2 \in \mathbb{R}.$ Thus, we have $a^2+4a+40+r^2-12r=0,$ and the discriminant of this must be nonnegative as $a$ is real. Thus, $16-4(40+r^2-12r) \geq 0$ or $(r-6)^2 \leq 0,$ which forces $r = 6,$ as claimed.

Thus, we plug in $z \overline{z} = 6,$ and get: $72 = 4(z + \overline{z}) + 76 + (z + \overline{z})^2,$ ie. $(z+\overline{z})^2 + 4(z + \overline{z}) + 4 = 0,$ or $(z+\overline{z} + 2)^2 = 0,$ which means $z + \overline{z} = \boxed{-2}$ and that's our answer since we know $\overline{z} = 6 / z$

- ccx09

Solution 5

Observe that all the answer choices are real. Therefore, $z$ and $\frac{6}{z}$ must be complex conjugates as this is the only way for both their sum (one of the answer choices) and their product ( $6$ ) to be real. Thus $|z|=|\tfrac{6}{z}|=\sqrt{6}$ . We will test all the answer choices, starting with $\textbf{(A)}$ . Suppose the answer is $\textbf{(A)}$ . If $z+\tfrac{6}{z}=-2$ then $z^{2}+2z+6=0$ and $z=\frac{-2\pm\sqrt{4-24}}{2}=-1\pm\sqrt{5}i$ . Note that if $z=-1+\sqrt{5}i$ works, then so does $-1-\sqrt{5}i$ . It is relatively easy to see that if $z=-1+\sqrt{5}i$ , then $12|z|^{2}=12\cdot 6=72, 2|z+2|^{2}=2|1+\sqrt{5}i|=2\cdot 6=12, |z^{2}+1|^{2}=|-3-2\sqrt{5}i|^{2}=29,$ and $72=12+29+31$ . Thus the condition $12|z|^{2}=2|z+2|^{2}+|z^{2}+1|^{2}+31$ is satisfied for $z+\tfrac{6}{z}=-2$ , and the answer is $\boxed{\textbf{(A)} ~-2}$ .

Video Solution by Punxsutawney Phil

https://youtu.be/E0HkYqZzw3s

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