2009 AIME I Problems/Problem 2

Problem

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 $n = \boxed{697}.$