AMC12 2023 B

AMC12 2023 B · Q12

AMC12 2023 B · Q12. It mainly tests Linear equations, Complex numbers (rare).

For complex number $u = a+bi$ and $v = c+di$ (where $i=\sqrt{-1}$), define the binary operation $u \otimes v = ac + bdi$ Suppose $z$ is a complex number such that $z\otimes z = z^{2}+40$. What is $|z|$?

对于复数$u = a+bi$和$v = c+di$（其中$i=\sqrt{-1}$），定义二元运算 $u \otimes v = ac + bdi$ 假设$z$是一个复数使得$z\otimes z = z^{2}+40$。$|z|$是多少？

(A) 2 2

(B) 5 5

(D) \sqrt{10} \sqrt{10}

(E) 5\sqrt{2} 5\sqrt{2}

Answer

Correct choice: (E)

正确答案：（E）

Solution

let $z$ = $a+bi$. $z \otimes z = a^{2}+b^{2}i$. This is equal to $z^{2} + 40 = a^{2}-b^{2}+40+2abi$ Since the real values have to be equal to each other, $a^{2}-b^{2}+40 = a^{2}$. Simple algebra shows $b^{2} = 40$, so $b = \pm 2\sqrt{10}$. The imaginary components must also equal each other, meaning $b^{2} = 2ab$, or $b = 2a$. This means $a = \frac{b}{2} = \pm \sqrt{10}$. Thus, the magnitude of $z$ is $\sqrt{a^{2}+b^{2}} = \sqrt{10+40} = \sqrt{50} = 5\sqrt{2}$ $=\text{\boxed{\textbf{(E) }5\sqrt{2}}}$

设$z = a+bi$。 $z \otimes z = a^{2}+b^{2}i$。这等于$z^{2} + 40 = a^{2}-b^{2}+40+2abi$。实部必须相等，因此$a^{2}-b^{2}+40 = a^{2}$。简单代数得$b^{2} = 40$，所以$b = \pm 2\sqrt{10}$。虚部也必须相等，即$b^{2} = 2ab$，或$b = 2a$。因此$a = \frac{b}{2} = \pm \sqrt{10}$。于是$z$的模为$\sqrt{a^{2}+b^{2}} = \sqrt{10+40} = \sqrt{50} = 5\sqrt{2}$ $=\boxed{\textbf{(E) }5\sqrt{2}}$。

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