# 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, $$\frac {4i}{17}=\frac {164i}{n}$$ $$n=\frac {(164)(17)}{4}=697$$ $$n = \boxed{697}.$$

## Solution 3 (Highly not recommended)

Below is an image of the complex plane. Let $\operatorname{Im}(z)$ denote the imaginary part of a complex number $z$. [asy] unitsize(1cm); xaxis("Re",Arrows); yaxis("Im",Arrows); real f(real x) {return 164;} pair F(real x) {return (x,f(x));} draw(graph(f,-700,100),red,Arrows); label("Im(z)=164",F(-330),N); pair z = (-656,164); dot(Label("z",align=N),z); dot(Label("z+n",align=N),z+(697,0)); draw(Label("4x"),z--(0,0)); draw(Label("x"),(0,0)--z+(697,0)); markscalefactor=2; draw(rightanglemark(z,(0,0),z+(697,0))); [/asy] $z$ must lie on the line $\operatorname{Im}(z)=164$. $z+n$ must also lie on the same line, since $n$ is real and does not affect the imaginary part of $z$.

Consider $z$ and $z+n$ in terms of their magnitude (distance from the origin) and phase (angle formed by the point, the origin, and the positive real axis, measured counterclockwise from said axis). When multiplying/dividing two complex numbers, you can multiply/divide their magnitudes and add/subtract their phases to get the magnitude and phase of the product/quotient. Expressed in a formula, we have $z_1z_2 = r_1\angle\theta_1 \cdot r_2\angle\theta_2 = r_1r_2\angle(\theta_1+\theta_2)$ and $\frac{z_1}{z_2} = \frac{r_1\angle\theta_1}{r_2\angle\theta_2} = \frac{r_1}{r_2}\angle(\theta_1-\theta_2)$, where $r$ is the magnitude and $\theta$ is the phase, and $z_n=r_n\angle\theta_n$.

Since $4i$ has magnitude $4$ and phase $90^\circ$ (since the positive imaginary axis points in a direction $90^\circ$ counterclockwise from the positive real axis), $z$ must have a magnitude $4$ times that of $z+n$. We denote the length from the origin to $z+n$ with the value $x$ and the length from the origin to $z$ with the value $4x$. Additionally, $z$, the origin, and $z+n$ must form a right angle, with $z$ counterclockwise from $z+n$.

This means that $z$, the origin, and $z+n$ form a right triangle. The hypotenuse is the length from $z$ to $z+n$ and has length $n$, since $n$ is defined to be a positive integer. The area of the triangle can be expressed using the two legs, as $\frac{x \cdot 4x}{2}$, or using the hypotenuse and its corresponding altitude, as $\frac{164n}{2}$, so $\frac{x \cdot 4x}{2} = \frac{164n}{2} \implies x^2 = 41n$. By Pythagorean Theorem, $x^2+(4x)^2 = n^2 \implies 17x^2 = n^2$. Substituting out $x^2$ using the earlier equation, we get $17\cdot41n = n^2 \implies n = \boxed{697}$. ~emerald_block

~IceMatrix

~Shreyas S

## See also

 2009 AIME I (Problems • Answer Key • Resources) Preceded byProblem 1 Followed byProblem 3 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions

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