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

Video Solution

https://youtu.be/NL79UexadzE

~IceMatrix

Video Solution

https://www.youtube.com/watch?v=P00iOJdQiL4

~Shreyas S

See also

2009 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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All AIME Problems and Solutions

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