2009 AMC 12A Problems/Problem 15
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Problem
For what value of 
is 
?
Note: here 
.
Solution
Obviously, even powers of 
are real and odd powers of are imaginary.
Hence the real part of the sum is 
, and
the imaginary part is 
.
Let's take a look at the real part first. We have 
, hence the real part simplifies to 
.
If there were an odd number of terms, we could pair them as follows: 
, hence the result would be negative. As we need the real part to be 
, we must have an even number of terms. If we have an even number of terms, we can pair them as 
. Each parenthesis is equal to 
, thus there are 
of them, and the last value used is 
. This happens for 
and 
. As is not present as an option, we may conclude that the answer is 
.
In a complete solution, we should now verify which of and will give us the correct imaginary part.
We can rewrite the imaginary part as follows: 
. We need to obtain 
. Once again we can repeat the same reasoning: If the number of terms were even, the left hand side would be negative, thus the number of terms is odd. The left hand side can then be rewritten as 
. We need parentheses, therefore the last value used is 
. This happens when or 
, and we are done.
See Also
| 2009 AMC 12A (Problems • Resources) | ||
| Preceded by Problem 14 | Followed by Problem 16 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
