Difference between revisions of "2019 AMC 12A Problems/Problem 21"
(Created page with "==Problem== Let <cmath>z=\frac{1+i}{\sqrt{2}}.</cmath>What is <cmath>(z^{1^2}+z^{2^2}+z^{3^2}+\dots+z^{{12}^2}) \cdot (\frac{1}{z^{1^2}}+\frac{1}{z^{2^2}}+\frac{1}{z^{3^2}}+\...") |
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==Solution== | ==Solution== | ||
+ | |||
+ | Note that <math>z = \mathrm{cis }(45)</math>. | ||
+ | |||
+ | Also note that <math>z^{k} = z^{k + 8}</math> for all positive integers <math>k</math> because of DeMoivre's Theorem. Therefore, we want to look at the exponents of each term mod 8. | ||
+ | |||
+ | <math>1^2, 5^2,</math> and <math>9^2</math> are all <math>1 \pmod{8}</math> | ||
+ | |||
+ | <math>2^2, 6^2,</math> and <math>10^2</math> are all <math>4 \pmod{8}</math> | ||
+ | |||
+ | <math>3^2, 7^2,</math> and <math>11^2</math> are all <math>1 \pmod{8}</math> | ||
+ | |||
+ | <math>4^2, 8^2,</math> and <math>12^2</math> are all <math>0 \pmod{8}</math> | ||
+ | |||
+ | Therefore, | ||
+ | |||
+ | <math>z^{1^2} = z^{5^2} = z^{9^2} = \mathrm{cis }(45)</math> | ||
+ | |||
+ | <math>z^{2^2} = z^{6^2} = z^{10^2} = \mathrm{cis }(180) = -1</math> | ||
+ | |||
+ | <math>z^{3^2} = z^{7^2} = z^{11^2} = \mathrm{cis }(45)</math> | ||
+ | |||
+ | <math>z^{4^2} = z^{8^2} = z^{12^2} = \mathrm{cis }(0) = 1</math> | ||
+ | |||
+ | The term <math>(z^{1^2}+z^{2^2}+z^{3^2}+\dots+z^{{12}^2})</math> simplifies to <math>6\mathrm{cis }(45)</math>, while the term <math>(\frac{1}{z^{1^2}}+\frac{1}{z^{2^2}}+\frac{1}{z^{3^2}}+\dots+\frac{1}{z^{{12}^2}})</math> simplifies to <math>\frac{6}{\mathrm{cis }(45)}</math>. Upon multiplication, the <math>\mathrm{cis }(45)</math> cancels out and leaves us with <math>\boxed{\textbf{(C) }36}</math>. | ||
==See Also== | ==See Also== |
Revision as of 17:57, 9 February 2019
Problem
Let What is
Solution
Note that .
Also note that for all positive integers
because of DeMoivre's Theorem. Therefore, we want to look at the exponents of each term mod 8.
and
are all
and
are all
and
are all
and
are all
Therefore,
The term simplifies to
, while the term
simplifies to
. Upon multiplication, the
cancels out and leaves us with
.
See Also
2019 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
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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