Difference between revisions of "2017 AMC 12A Problems/Problem 17"
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By [[Euler's identity]], <math>1 = e^{0 \cdot i} = cos (2k\pi) + i sin(2k\pi)</math>, where <math>k</math> is an integer. | By [[Euler's identity]], <math>1 = e^{0 \cdot i} = cos (2k\pi) + i sin(2k\pi)</math>, where <math>k</math> is an integer. | ||
− | Using [[De Moivre's Theorem]], we have <math>z = 1^{\frac{1}{24}} = {cos (\frac{k\pi}{12}) + i sin (\frac{k\pi}{12})}</math>, where <math>0 \leq k<24</math>. | + | Using [[De Moivre's Theorem]], we have <math>z = 1^{\frac{1}{24}} = {cos (\frac{k\pi}{12}) + i sin (\frac{k\pi}{12})}</math>, where <math>0 \leq k<24</math> that produce <math>24</math> unique results. |
Using De Moivre's Theorem again, we have <math>z^6 = {cos (\frac{k\pi}{2}) + i sin (\frac{k\pi}{2})}</math> | Using De Moivre's Theorem again, we have <math>z^6 = {cos (\frac{k\pi}{2}) + i sin (\frac{k\pi}{2})}</math> |
Revision as of 02:12, 9 February 2017
Contents
Problem
There are different complex numbers
such that
. For how many of these is
a real number?
Solution
Note that these such that
are
for integer
. So
This is real iff is even
. Thus, the answer is the number of even
which is
.
Solution 2
By Euler's identity, , where
is an integer.
Using De Moivre's Theorem, we have , where
that produce
unique results.
Using De Moivre's Theorem again, we have
For to be real,
has to equal
to negate the imaginary component. This occurs whenever
is an integer multiple of
, requiring that
is even. There are exactly
even values of
on the interval
, so the answer is
.
See Also
2017 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
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 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.