# 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 $n$.

## Solution 1

Let $z = a + 164i$.

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

By comparing coefficients, equating the real terms on the leftmost and rightmost side of the equation,

we conclude that $$a = -656.$$

By equating the imaginary terms on each side of the equation,

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

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

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

## Solution 2

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

$$1-\frac {n}{z+n}=4i$$

$$1-4i=\frac {n}{z+n}$$

$$\frac {1}{1-4i}=\frac {z+n}{n}$$

$$\frac {1+4i}{17}=\frac {z}{n}+1$$

Since their imaginary part has to be equal,

$$\frac {4i}{17}=\frac {164i}{n}$$

$$n=\frac {(164)(17)}{4}=697$$

$$n = \boxed{697}.$$