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 | |
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