2009 AIME I Problems/Problem 2: Difference between revisions

Revision as of 19:31, 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 $z$ = $a$ + $164$ $i$ .

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 9: / Line 9: @@
 == Solution ==
-Let <math>z</math> = <math>a</math> + <math>b</math><math>i</math>
+Let <math>z</math> = <math>a</math> + <math>164</math><math>i</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>
+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>
+and this equation shows that <math>n = \boxed{697}.</math>