1988 AHSME Problems/Problem 21
Problem
The complex number satisfies . What is ? Note: if , then .
Solution
Let the complex number equal . Then the preceding equation can be expressed as Because and must both be real numbers, we immediately have that , giving . Plugging this in back to our equation gives us . Rearranging this into , we can square each side of the equation resulting in Further simplification will yield meaning that . Knowing both and , we can plug them in into . Our final answer is .
See also
1988 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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