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 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 has 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, the smallest integer greater is
~ PaperMath ~megaboy6679
Solution 4
First, let where and are real numbers. We now have that given the conditions 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, it is , and our answer is .
Now, if then we have that . Therefore, is not 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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