Difference between revisions of "2009 AIME I Problems/Problem 2"
Line 9: | Line 9: | ||
== Solution == | == Solution == | ||
− | Let <math>z | + | Let <math>z = a + 164i</math>. |
− | Then <cmath>\frac {a + 164i}{a + 164i + n} = 4i</cmath> and <cmath>a + 164i = \left (4i \right ) \left (a + n + 164i \right ) = 4i \left (a + n \right ) - 656.</cmath> | + | Then |
+ | <cmath>\frac {a + 164i}{a + 164i + n} = 4i</cmath> | ||
+ | and | ||
+ | <cmath>a + 164i = \left (4i \right ) \left (a + n + 164i \right ) = 4i \left (a + n \right ) - 656.</cmath> | ||
− | From this, we conclude that <cmath>a = -656</cmath> and <cmath>164i = 4i \left (a + n \right ) = 4i \left (-656 + n \right ).</cmath> | + | From this, we conclude that |
+ | <cmath>a = -656</cmath> | ||
+ | and | ||
+ | <cmath>164i = 4i \left (a + n \right ) = 4i \left (-656 + n \right ).</cmath> | ||
We now have an equation for <math>n</math>: <cmath>4i \left (-656 + n \right ) = 164i,</cmath> | We now have an equation for <math>n</math>: <cmath>4i \left (-656 + n \right ) = 164i,</cmath> | ||
and this equation shows that <math>n = \boxed{697}.</math> | and this equation shows that <math>n = \boxed{697}.</math> |
Revision as of 19:33, 19 March 2009
Problem
There is a complex number with imaginary part and a positive integer such that
Find .
Solution
Let .
Then and
From this, we conclude that and
We now have an equation for :
and this equation shows that