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

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For complex number <math>u = a+bi</math> and <math>v = c+di</math> (where <math>i=\sqrt{-1}</math>), define the binary operation
 
For complex number <math>u = a+bi</math> and <math>v = c+di</math> (where <math>i=\sqrt{-1}</math>), define the binary operation
  
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<math>u \earth v = ac \sum bdi</math>
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<math>u \otimes v = ac + bdi</math>
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<math>\textbf{(A) }\frac{9}{25}\qquad\textbf{(B) }\frac{1}{9}\qquad\textbf{(C) }\frac{1}{5}\qquad\textbf{(D) }\frac{25}{169}\qquad\textbf{(E) }\frac{4}{25}</math>
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Suppose <math>z</math> is a complex number such that <math>z\otimes z = z^{2}+40</math>. What is <math>|z|</math>?
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<math>\textbf{(A) }2\qquad\textbf{(B) }5\qquad\textbf{(C) }\sqrt{5}\qquad\textbf{(D) }\sqrt{10}\qquad\textbf{(E) }5\sqrt{2}</math>
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==Solution 1==
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let <math>z</math> = <math>a+bi</math>.
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<math>z \otimes z = a^{2}+b^{2}i</math>.
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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>.
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Simple algebra shows <math>b^{2} = 40</math>, so <math>b</math> is <math>2\sqrt{10}</math>.
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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>.
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Thus, the magnitude of z is <math> \sqrt{a^{2}+b^{2}} = \sqrt{50} = 5\sqrt{2}</math>
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<math>=\text{\boxed{\textbf{(E) }5\sqrt{2}}}</math>
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~Failure.net
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==Video Solution==
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https://youtu.be/Yw3W2ptPWYQ
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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==See Also==
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{{AMC12 box|year=2023|ab=B|num-b=11|num-a=13}}
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{{MAA Notice}}

Latest revision as of 01:52, 20 November 2023

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 $b$ 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

Video Solution

https://youtu.be/Yw3W2ptPWYQ


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

See Also

2023 AMC 12B (Problems • Answer Key • Resources)
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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