2004 AMC 12B Problems/Problem 16
Problem
A function is defined by , where and is the complex conjugate of . How many values of satisfy both and ?
Solutions
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
Solution 3
Let , like above. Therefore, . We move some terms around to get . We factor: . We divide out the common factor to see that . Next we put this into the definition of . Finally, , and has two solutions.
See also
2004 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 15 |
Followed by Problem 17 |
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All AMC 12 Problems and Solutions |
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