Difference between revisions of "2017 AMC 12A Problems/Problem 17"
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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> | ||
− | For <math>z^6</math> to be real, <math>sin(\frac{k\pi}{2})</math> has be equal <math>0</math> to negate the imaginary component. This occurs whenever <math>\frac{k\pi}{2}</math> is | + | For <math>z^6</math> to be real, <math>sin(\frac{k\pi}{2})</math> has be equal <math>0</math> to negate the imaginary component. This occurs whenever <math>\frac{k\pi}{2}</math> is an integer multiple of <math>\pi</math>, requiring that <math>k</math> is even. There are exactly <math>\boxed{12}</math> even values of <math>k</math> on the interval <math>0 \leq k<24</math>, so the answer is <math>\boxed{(D)}</math>. |
==See Also== | ==See Also== | ||
{{AMC12 box|year=2017|ab=A|num-b=16|num-a=18}} | {{AMC12 box|year=2017|ab=A|num-b=16|num-a=18}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 02:10, 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
.
Using De Moivre's Theorem again, we have
For to be real,
has be 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 |
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