2018 AIME II Problems/Problem 5
Contents
[hide]Problem
Suppose that , , and are complex numbers such that , , and , where . Then there are real numbers and such that . Find .
Solution 1
First we evaluate the magnitudes. , , and . Therefore, , or . Divide to find that , , and . This allows us to see that the argument of is , and the argument of is . We need to convert the polar form to a standard form. Simple trig identities show and . More division is needed to find what is. Written by a1b2
Solution 2
Dividing the first equation by the second equation given, we find that . Substituting this into the third equation, we get . Taking the square root of this is equivalent to halving the argument and taking the square root of the magnitude. Furthermore, the second equation given tells us that the argument of is the negative of that of , and their magnitudes multiply to . Thus we have and . To find , we can use the previous substitution we made to find that Therefore, Solution by ktong
2018 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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Solution 3
We are given that . Thus . We are also given that . Thus . We are also given that = . Substitute and into = . We have . Multiplying out we get . Thus . Simplifying this fraction we get . Cross-multiplying the fractions we get or . Now we can rewrite this as . Let .Thus .