Difference between revisions of "2004 AMC 12B Problems/Problem 16"
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\qquad\mathrm{(E)}\ 8</math> | \qquad\mathrm{(E)}\ 8</math> | ||
− | == Solution == | + | == Solution 1== |
− | Let <math>z = a+bi</math>, so <math>\overline{z} = a-bi</math>. By definition, <math>z = a+bi = f(z) = i(a-bi) = b+ai</math>, which implies that all solutions to <math>f(z) = z</math> lie on the line <math>y=x</math> on the complex plane. The graph of <math>|z| = 5</math> is a [[circle]] centered at the origin, and there are <math>2 \Rightarrow \mathrm{(C)}</math> intersections. | + | Let <math>z = a+bi</math>, so <math>\overline{z} = a-bi</math>. By definition, <math>z = a+bi = f(z) = i(a-bi) = b+ai</math>, which implies that all solutions to <math>f(z) = z</math> lie on the line <math>y=x</math> on the complex plane. The graph of <math>|z| = 5</math> is a [[circle]] centered at the origin, and there are <math>2 \Rightarrow \mathrm{(C)}</math> intersections. |
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+ | ===Solution 2=== | ||
+ | We start the same as the above solution: Let <math>z = a+bi</math>, so <math>\overline{z} = a-bi</math>. By definition, <math>z = a+bi = f(z) = i(a-bi) = b+ai</math>. Since we are given <math>|z| = 5</math>, this implies that <math>a^2+b^2=25</math>. We recognize the Pythagorean triple <math>3,4,5</math> so we see that <math>(a,b)=(3,4)</math> or <math>(4,3)</math>. So the answer is <math>2 \Rightarrow \mathrm{(C)}</math>. | ||
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+ | Solution by franzliszt | ||
== See also == | == See also == |
Revision as of 08:47, 8 July 2020
Contents
Problem
A function is defined by , where and is the complex conjugate of . How many values of satisfy both and ?
Solution 1
Let , so . By definition, , which implies that all solutions to lie on the line on the complex plane. The graph of is a circle centered at the origin, and there are intersections.
Solution 2
We start the same as the above solution: Let , so . By definition, . Since we are given , this implies that . We recognize the Pythagorean triple so we see that or . So the answer is .
Solution by franzliszt
See also
2004 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 15 |
Followed by Problem 17 |
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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