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2023 AMC 12B Problems/Problem 12: Difference between revisions

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m Solution 1: a and b can be negative, though this doesn't affect the final answer
 
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Since the real values have to be equal to each other, <math>a^{2}-b^{2}+40 = a^{2}</math>.
Since the real values have to be equal to each other, <math>a^{2}-b^{2}+40 = a^{2}</math>.
Simple algebra shows <math>b^{2} = 40</math>, so <math>b</math> is <math>2\sqrt{10}</math>.
Simple algebra shows <math>b^{2} = 40</math>, so <math>b = \pm 2\sqrt{10}</math>.


The imaginary components must also equal each other, meaning <math>b^{2} = 2ab</math>, or <math>b = 2a</math>. This means <math>a = \frac{b}{2} = \sqrt{10}</math>.
The imaginary components must also equal each other, meaning <math>b^{2} = 2ab</math>, or <math>b = 2a</math>. This means <math>a = \frac{b}{2} = \pm \sqrt{10}</math>.


Thus, the magnitude of z is <math> \sqrt{a^{2}+b^{2}} = \sqrt{50} = 5\sqrt{2}</math>
Thus, the magnitude of <math>z</math> is <math> \sqrt{a^{2}+b^{2}} = \sqrt{10+40} = \sqrt{50} = 5\sqrt{2}</math>
<math>=\text{\boxed{\textbf{(E) }5\sqrt{2}}}</math>
<math>=\text{\boxed{\textbf{(E) }5\sqrt{2}}}</math>



Latest revision as of 16:41, 24 October 2025

Problem

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 $|z|$?

$\textbf{(A) }2\qquad\textbf{(B) }5\qquad\textbf{(C) }\sqrt{5}\qquad\textbf{(D) }\sqrt{10}\qquad\textbf{(E) }5\sqrt{2}$

Solution 1

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 $\sqrt{a^{2}+b^{2}} = \sqrt{10+40} = \sqrt{50} = 5\sqrt{2}$ $=\text{\boxed{\textbf{(E) }5\sqrt{2}}}$

~Failure.net

Video Solution

https://youtu.be/Yw3W2ptPWYQ


~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

See Also

2023 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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