Difference between revisions of "2002 AIME I Problems/Problem 12"

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== Problem ==
 
== Problem ==
Let <math>F(z)=\dfrac{z+i}{z-i}</math> for all complex numbers <math>z\neq 1</math>, and let <math>z_n=F(z_{n-1})</math> for all positive integers <math>n</math>. Given that <math>z_0=\dfrac{1}{137}+i</math> and <math>z_{2002}=a+bi</math>, where <math>a</math> and <math>b</math> are real numbers, find <math>a+b</math>.
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Let <math>F(z)=\dfrac{z+i}{z-i}</math> for all complex numbers <math>z\neq i</math>, and let <math>z_n=F(z_{n-1})</math> for all positive integers <math>n</math>. Given that <math>z_0=\dfrac{1}{137}+i</math> and <math>z_{2002}=a+bi</math>, where <math>a</math> and <math>b</math> are real numbers, find <math>a+b</math>.
  
 
== Solution ==
 
== Solution ==

Revision as of 21:57, 4 May 2008

Problem

Let $F(z)=\dfrac{z+i}{z-i}$ for all complex numbers $z\neq i$, and let $z_n=F(z_{n-1})$ for all positive integers $n$. Given that $z_0=\dfrac{1}{137}+i$ and $z_{2002}=a+bi$, where $a$ and $b$ are real numbers, find $a+b$.

Solution

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See also

2002 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions
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