Difference between revisions of "2017 AMC 12A Problems/Problem 17"
m (→Solution 1) |
|||
Line 26: | Line 26: | ||
==Solution 3== | ==Solution 3== | ||
From the start, recall from the Fundamental Theorem of Algebra that <math>z^{24} = 1</math> must have <math>24</math> solutions (and these must be distinct since the equation factors into <math>0 = (z-1)(z^{23} + z^{22} + z^{21}... + z + 1)</math>), or notice that the question is simply referring to the 24th roots of unity, of which we know there must be <math>24</math>. Notice that <math>1 = z^{24} = (z^6)^4</math>, so for any solution <math>z</math>, <math>z^6</math> will be one of the 4th roots of unity (<math>1</math>, <math>i</math>, <math>-1</math>, or <math>-i</math>). Then <math>6</math> solutions <math>z</math> will satisfy <math>z^6 = 1</math>, <math>6</math> will satisfy <math>z^6 = -1</math> (and this is further justified by knowledge of the 6th roots of unity), so there must be <math>\boxed{(D) \: 12}</math> such <math>z</math>. | From the start, recall from the Fundamental Theorem of Algebra that <math>z^{24} = 1</math> must have <math>24</math> solutions (and these must be distinct since the equation factors into <math>0 = (z-1)(z^{23} + z^{22} + z^{21}... + z + 1)</math>), or notice that the question is simply referring to the 24th roots of unity, of which we know there must be <math>24</math>. Notice that <math>1 = z^{24} = (z^6)^4</math>, so for any solution <math>z</math>, <math>z^6</math> will be one of the 4th roots of unity (<math>1</math>, <math>i</math>, <math>-1</math>, or <math>-i</math>). Then <math>6</math> solutions <math>z</math> will satisfy <math>z^6 = 1</math>, <math>6</math> will satisfy <math>z^6 = -1</math> (and this is further justified by knowledge of the 6th roots of unity), so there must be <math>\boxed{(D) \: 12}</math> such <math>z</math>. | ||
+ | |||
+ | ==Solution 4 (Quick)== | ||
+ | Let <math>a\in\mathbb{R}</math> and <math>a = z^6.</math> We have | ||
+ | <cmath>a^4 = 1 \implies a = 1,-1.</cmath> | ||
+ | <math>z^6 = \pm 1</math> has 6 solutions for <math>1</math> and <math>-1</math> respectively, so <math>6+6=\boxed{(D)\ 12}.</math> <cmath> </cmath> | ||
+ | -svyn | ||
==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}} |
Latest revision as of 04:19, 2 February 2021
Problem
There are different complex numbers such that . For how many of these is a real number?
Solution 1
Note that these such that are for integer . So
This is real if 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 .
Solution 3
From the start, recall from the Fundamental Theorem of Algebra that must have solutions (and these must be distinct since the equation factors into ), or notice that the question is simply referring to the 24th roots of unity, of which we know there must be . Notice that , so for any solution , will be one of the 4th roots of unity (, , , or ). Then solutions will satisfy , will satisfy (and this is further justified by knowledge of the 6th roots of unity), so there must be such .
Solution 4 (Quick)
Let and We have has 6 solutions for and respectively, so -svyn
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.