2002 AMC 12A Problems/Problem 24
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Problem
Find the number of ordered pairs of real numbers such that .
Solution
Let be the magnitude of . Then the magnitude of is , while the magnitude of is . We get that , hence either or .
For we get a single solution .
Let's now assume that . Multiply both sides by . The left hand side becomes , the right hand side becomes . Hence the solutions for this case are precisely all the rd complex roots of unity, and there are of those.
The total number of solutions is therefore .
See Also
2002 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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All AMC 12 Problems and Solutions |