AoPSWiki

Art of Problem Solving celebrates the many
accomplishments of its students and community members.

Personal tools

Search

Toolbox

2009 AIME I Problems/Problem 2

From AoPSWiki

Revision as of 04:45, 28 March 2009 by Boy Soprano II (Talk | contribs)

(diff) ← Older revision | Current revision (diff) | Newer revision → (diff)

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

1st Solution

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 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 $n = \boxed{697}.$

2nd Solution

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

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

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

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

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

Since their imaginery part has to be equal,

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

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

$n = \boxed{697}.$

See also

2009 AIME I (Problems • Resources)
Preceded by Problem 1	Followed by Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2009_AIME_I_Problems/Problem_2"

Category: Complex numbers

Want to learn how to tackle those tough AMC/AIME/Olympiad counting and probability problems? Check out Art of Problem Solving's Intermediate Counting & Probability by David Patrick.

© Copyright 2008 AoPS Incorporated. All Rights Reserved. • Foundation • Privacy • Contact Us