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Difference between revisions of "2023 AMC 12B Problems/Problem 12"

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This is equal to <math>z^{2} + 40 = a^{2}-b^{2}+40+2abi</math>
 
This is equal to <math>z^{2} + 40 = a^{2}-b^{2}+40+2abi</math>
  
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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</math> is <math>2\sqrt{10}</math>.
  

Revision as of 17:01, 15 November 2023

Problem

For complex number $u = a+bi$ and $v = c+di$ (where $i=\sqrt{-1}$), define the binary operation

$u \odot v = ac + bdi$

Suppose $z$ is a complex number such that $z\odot z = z^{2}+40$. 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 $z$ = $a+bi$.

$z \odot z = a^{2}+b^{2}i$.

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 $b$ is $2\sqrt{10}$.

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

Thus, the magnitude of z is $\sqrt{a^{2}+b^{2}} = \sqrt{50} = 5\sqrt{2}$ $=\text{\boxed{\textbf{(E) }5\sqrt{2}}}$

~Failure.net

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