Difference between revisions of "2002 AMC 12A Problems/Problem 24"
(New page: == Problem == Find the number of ordered pairs of real numbers <math>(a,b)</math> such that <math>(a+bi)^{2002} = a-bi</math>. <math> \text{(A) }1001 \qquad \text{(B) }1002 \qquad \text{...) |
m (Corrected triviality 2003th --> 2003rd) |
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Line 21: | Line 21: | ||
For <math>s=0</math> we get a single solution <math>(a,b)=(0,0)</math>. | For <math>s=0</math> we get a single solution <math>(a,b)=(0,0)</math>. | ||
− | Let's now assume that <math>s=1</math>. Multiply both sides by <math>a+bi</math>. The left hand side becomes <math>(a+bi)^{2003}</math>, the right hand side becomes <math>(a-bi)(a+bi)=a^2 + b^2 = 1</math>. Hence the solutions for this case are precisely all the <math>2003</math> | + | Let's now assume that <math>s=1</math>. Multiply both sides by <math>a+bi</math>. The left hand side becomes <math>(a+bi)^{2003}</math>, the right hand side becomes <math>(a-bi)(a+bi)=a^2 + b^2 = 1</math>. Hence the solutions for this case are precisely all the <math>2003</math>rd complex roots of unity, and there are <math>2003</math> of those. |
The total number of solutions is therefore <math>1+2003 = \boxed{2004}</math>. | The total number of solutions is therefore <math>1+2003 = \boxed{2004}</math>. |
Revision as of 01:19, 27 June 2010
Problem
Find the number of ordered pairs of real numbers such that .
Solution
Let be the magnitude of . Then the magnitude of is , while the magnitude of is . We get that , hence either or .
For we get a single solution .
Let's now assume that . Multiply both sides by . The left hand side becomes , the right hand side becomes . Hence the solutions for this case are precisely all the rd complex roots of unity, and there are of those.
The total number of solutions is therefore .
See Also
2002 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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 |