# Difference between revisions of "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, 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 $x$ is real and $x^2<24$, so $z = \frac{x \pm i\sqrt{24-x^2}}{2}$. We'll take the plus sign for now since we know the answer is unique. Then we have $$|z|^2 = 6$$ $$|z+2|^2 = (\frac{x}{2} + 2)^2 + \frac{24-x^2}{4} = 2x+10$$ $$|z^2+1|^2 = |xz -6 +1|^2 = (\frac{x^2}{2}-5)^2 + \frac{x^2(24-x^2)}{4} = x^2 +25$$ Plug the above back to the original equation, we have $$12*6 = 2(2x+10) + x^2 + 25 + 31$$ $$(x+2)^2 = 0$$ So $x = -2$ $\boxed{\textbf{(A) }-2}$.

~Sequoia

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 