Difference between revisions of "2023 AMC 12B Problems/Problem 12"
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==Solution 1== | ==Solution 1== | ||
let <math>z</math> = <math>a+bi</math>. | let <math>z</math> = <math>a+bi</math>. | ||
− | <math>z \odot z a^{2}+b^{2}i</math>. | + | |
+ | <math>z \odot z = a^{2}+b^{2}i</math>. | ||
+ | |||
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> | ||
+ | |||
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>. | ||
+ | |||
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} = \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 z is <math> \sqrt{a^{2}+b^{2}} = \sqrt{50} = 5\sqrt{2}</math> | ||
<math>=\text{\boxed{\textbf{(E) }5\sqrt{2}}}</math> | <math>=\text{\boxed{\textbf{(E) }5\sqrt{2}}}</math> | ||
~Failure.net | ~Failure.net |
Revision as of 17:01, 15 November 2023
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 is .
The imaginary components must also equal each other, meaning , or . This means .
Thus, the magnitude of z is
~Failure.net