AMC12 2021 B

AMC12 2021 B · Q18

AMC12 2021 B · Q18. It mainly tests Complex numbers (rare).

Let $z$ be a complex number satisfying $12|z|^2=2|z+2|^2+|z^2+1|^2+31.$ What is the value of $z+\frac 6z?$

设 $z$ 是一个满足 $12|z|^2=2|z+2|^2+|z^2+1|^2+31$ 的复数。$z+\frac 6z$ 的值为多少？

(A) -2 -2

(B) -1 -1

(D) 1 1

(E) 4 4

Answer

Correct choice: (A)

正确答案：（A）

Solution

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 \[z+\frac6z=z+\bar{z}=\boxed{\textbf{(A) }-2}.\]

利用 $z\bar{z}=|z|^2$，方程改写为 \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*} 由于括号内的两个量都是实数，二者必须都等于 $0$，因此 \[z+\frac6z=z+\bar{z}=\boxed{\textbf{(A) }-2}.\]

Topics

AL.019 Complex numbers (rare)