Difference between revisions of "2009 AIME I Problems/Problem 2"

Revision as of 20:39, 19 March 2009

There is a complex number $z$ with imaginary part $164$ and a positive integer $n$ such that

$\frac {z}{z + n} = 4i.$

Find $n$ .

Let $z = a + 164i$ .

Then $\frac {a + 164i}{a + 164i + n} = 4i$ and $a + 164i = \left (4i \right ) \left (a + n + 164i \right ) = 4i \left (a + n \right ) - 656.$

From this, we conclude that $a = -656$ and $164i = 4i \left (a + n \right ) = 4i \left (-656 + n \right ).$

We now have an equation for $n$ : $4i \left (-656 + n \right ) = 164i,$

and this equation shows that $n = \boxed{697}.$

@@ Line 24: / Line 24: @@
 and this equation shows that <math>n = \boxed{697}.</math>
+== See also ==
+{{AIME box|year=2009|n=I|num-b=1|num-a=3}}
+[[Category:Complex Numbers]]

2009 AIME I (Problems • Answer Key • Resources)
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