2000 AIME II Problems/Problem 9
Problem
Given that is a complex number such that , find the least integer that is greater than .
Solution
Using the quadratic equation on , we have .
There are other ways we can come to this conclusion. Note that if is on the unit circle in the complex plane, then and . We have and . Alternatively, we could let and solve to get .
Using De Moivre's Theorem we have , , so
.
We want .
Finally, the least integer greater than is .
Solution 2
Let . Notice that we have
must be (or else if you take the magnitude would not be the same). Therefore, and plugging into the desired expression, we get . Therefore, the least integer greater is
~solution by williamgolly
Solution 3 Intuitive
For this solution, we assume that and have the same least integer greater than their solution. we have . Since , . If we square the equation , we get , or . is is less than , since is less than . if we square the equation again, we get . since is less than 2, is less than 4, and is less than 2. However is also less than . we can see that every time we square the equation, the right hand side gets smaller, and into the negatives. Since the smallest integer that is allowed as an answer is 0, thus smallest integer greater is
~ PaperMath
Solution 4
First, let where and are real numbers. We now have that given the coniditons of the problem. Equating imaginary coefficients, we have that giving us that either or . Let's consider the latter case for now.
We now know that , so when we equate real coefficients we have that , therefore . So, and then we can write
By De Moivre's Theorem, . The imaginary parts cancel, leaving us with which is . Therefore, we have it being and our answer is .
Now, if then we have that . Therefore, is not real violating our conditions set above.
See also
2000 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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