Difference between revisions of "2009 AIME I Problems/Problem 2"
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<cmath>a + 164i = \left (4i \right ) \left (a + n + 164i \right ) = 4i \left (a + n \right ) - 656.</cmath> | <cmath>a + 164i = \left (4i \right ) \left (a + n + 164i \right ) = 4i \left (a + n \right ) - 656.</cmath> | ||
− | From this, we conclude that | + | From this, we conclude that (HOW?) |
<cmath>a = -656</cmath> | <cmath>a = -656</cmath> | ||
and | and |
Revision as of 21:24, 2 September 2011
Problem
There is a complex number with imaginary part
and a positive integer
such that
Find .
Solution
1st Solution
Let .
Then
and
From this, we conclude that (HOW?)
and
We now have an equation for :
and this equation shows that
2nd Solution
Since their imaginary part has to be equal,
See also
2009 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |