Difference between revisions of "1984 AHSME Problems/Problem 30"
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For any [[complex number]] <math> w=a+bi </math>, <math> |w| </math> is defined to be the [[real number]] <math> \sqrt{a^2+b^2} </math>. If <math> w=\cos40^\circ+i\sin40^\circ </math>, then <math> |w+2w^2+3w^3+...+9w^9|^{-1} </math> equals | For any [[complex number]] <math> w=a+bi </math>, <math> |w| </math> is defined to be the [[real number]] <math> \sqrt{a^2+b^2} </math>. If <math> w=\cos40^\circ+i\sin40^\circ </math>, then <math> |w+2w^2+3w^3+...+9w^9|^{-1} </math> equals | ||
− | <math> \ | + | <math> \text{(A) }\frac{1}{9}\sin40^\circ \qquad \text{(B) }\frac{2}{9}\sin20^\circ \qquad \text{(C) } \frac{1}{9}\cos40^\circ \qquad \text{(D) }\frac{1}{18}\cos20^\circ \qquad \text{(E) } \text{None of these} </math> |
==Solution== | ==Solution== | ||
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<cmath>S(1-w)=\sum_{i=1}^{9}\sum_{j=i}^{9} w^j(1-w)</cmath> | <cmath>S(1-w)=\sum_{i=1}^{9}\sum_{j=i}^{9} w^j(1-w)</cmath> | ||
− | + | By the geometric series formula, <math>\sum_{j=i}^{9} w^j(1-w)</math> is simply <math>w^i-w^{10}</math>. Therefore | |
<cmath>S(1-w)=\sum_{i=1}^{9} w^i-w^{10}=(w+w^2+\cdots +w^9)-9w^{10}</cmath> | <cmath>S(1-w)=\sum_{i=1}^{9} w^i-w^{10}=(w+w^2+\cdots +w^9)-9w^{10}</cmath> | ||
Latest revision as of 21:21, 17 July 2020
Problem
For any complex number ,
is defined to be the real number
. If
, then
equals
Solution
Let . Note that
Now we multiply by
:
By the geometric series formula, is simply
. Therefore
A simple application of De Moivre's Theorem shows that is a ninth root of unity (
), so
This shows that . Note that
, so
.
It's not hard to show that , so the number we seek is equal to
.
Now we plug into the fraction:
We multiply the numerator and denominator by and simplify to get
The absolute value of this is
Note that, from double angle formulas, , so
. Therefore
Therefore the correct answer is .
See Also
1984 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 29 |
Followed by Last Problem | |
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 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.