2023 AMC 12B Problems/Problem 12: Difference between revisions
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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 | ||
<math>u \ | <math>u \otimes v = ac + bdi</math> | ||
Suppose <math>z</math> is a complex number such that <math>z\ | Suppose <math>z</math> is a complex number such that <math>z\otimes z = z^{2}+40</math>. What is <math>|z|</math>? | ||
<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> | <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> | ||
==Solution 1== | |||
let <math>z</math> = <math>a+bi</math>. | |||
<math>z \otimes z = a^{2}+b^{2}i</math>. | |||
This is equal to <math>z^{2} + 40 = a^{2}-b^{2}+40+2abi</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 = \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} = \pm \sqrt{10}</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> | |||
~Failure.net | |||
==Video Solution== | |||
https://youtu.be/Yw3W2ptPWYQ | |||
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | |||
==See Also== | |||
{{AMC12 box|year=2023|ab=B|num-b=11|num-a=13}} | |||
{{MAA Notice}} | |||
Latest revision as of 16:41, 24 October 2025
Problem
For complex number
and
(where
), define the binary operation
Suppose
is a complex number such that
. What is
?
Solution 1
let
=
.
.
This is equal to
Since the real values have to be equal to each other,
.
Simple algebra shows
, so
.
The imaginary components must also equal each other, meaning
, or
. This means
.
Thus, the magnitude of
is
~Failure.net
Video Solution
~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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