Difference between revisions of "2009 AIME I Problems/Problem 2"

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== Solution ==
 
== Solution ==
  
Let <math>z</math> = <math>a</math> + <math>164</math><math>i</math>.
+
Let <math>z = a + 164i</math>.
  
Then <cmath>\frac {a + 164i}{a + 164i + n} = 4i</cmath> and <cmath>a + 164i = \left (4i \right ) \left (a + n + 164i \right ) = 4i \left (a + n \right ) - 656.</cmath>
+
Then
 +
<cmath>\frac {a + 164i}{a + 164i + n} = 4i</cmath>
 +
and
 +
<cmath>a + 164i = \left (4i \right ) \left (a + n + 164i \right ) = 4i \left (a + n \right ) - 656.</cmath>
  
From this, we conclude that <cmath>a = -656</cmath> and <cmath>164i = 4i \left (a + n \right ) = 4i \left (-656 + n \right ).</cmath>
+
From this, we conclude that
 +
<cmath>a = -656</cmath>
 +
and
 +
<cmath>164i = 4i \left (a + n \right ) = 4i \left (-656 + n \right ).</cmath>
  
 
We now have an equation for <math>n</math>: <cmath>4i \left (-656 + n \right ) = 164i,</cmath>
 
We now have an equation for <math>n</math>: <cmath>4i \left (-656 + n \right ) = 164i,</cmath>
  
 
and this equation shows that <math>n = \boxed{697}.</math>
 
and this equation shows that <math>n = \boxed{697}.</math>

Revision as of 19:33, 19 March 2009

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

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 $n = \boxed{697}.$