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

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== Problem ==
 
== Problem ==
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Let <math>F(z)=\dfrac{z+1}{z-1}</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 ==
−
{{solution}}
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Iterating <math>F</math> we get:
 +
 
 +
<cmath>
 +
\begin{align*}
 +
F(z) &= \frac{z+i}{z-i}\\
 +
F(F(z)) &= \frac{\frac{z+i}{z-i}+i}{\frac{z+i}{z-i}-i} = \frac{(z+i)+i(z-i)}{(z+i)-i(z-i)}= \frac{z+i+zi+1}{z+i-zi-1}= \frac{(z+1)(i+1)}{(z-1)(1-i)}\\
 +
&= \frac{(z+1)(i+1)^2}{(z-1)(1^2+1^2)}= \frac{(z+1)(2i)}{(z-1)(2)}= \frac{z+1}{z-1}i\\
 +
F(F(F(z))) &= \frac{\frac{z+1}{z-1}i+i}{\frac{z+1}{z-1}i-i} = \frac{\frac{z+1}{z-1}+1}{\frac{z+1}{z-1}-1} = \frac{(z+1)+(z-1)}{(z+1)-(z-1)} = \frac{2z}{2} = z.
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\end{align*}
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</cmath>
 +
 
 +
From this, it follows that <math>z_{k+3} = z_k</math>, for all <math>k</math>. Thus
 +
<math>z_{2002} = z_{3\cdot 667+1} = z_1 = \frac{z_0+i}{z_0-i} = \frac{(\frac{1}{137}+i)+i}{(\frac{1}{137}+i)-i}= \frac{\frac{1}{137}+2i}{\frac{1}{137}}= 1+274i.</math>
 +
 
 +
Thus <math>a+b = 1+274 = \boxed{275}</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2002|n=I|num-b=11|num-a=13}}
 
{{AIME box|year=2002|n=I|num-b=11|num-a=13}}
 +
{{MAA Notice}}

Latest revision as of 21:39, 21 November 2018

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

Iterating $F$ we get:

\begin{align*} F(z) &= \frac{z+i}{z-i}\\ F(F(z)) &= \frac{\frac{z+i}{z-i}+i}{\frac{z+i}{z-i}-i} = \frac{(z+i)+i(z-i)}{(z+i)-i(z-i)}= \frac{z+i+zi+1}{z+i-zi-1}= \frac{(z+1)(i+1)}{(z-1)(1-i)}\\ &= \frac{(z+1)(i+1)^2}{(z-1)(1^2+1^2)}= \frac{(z+1)(2i)}{(z-1)(2)}= \frac{z+1}{z-1}i\\ F(F(F(z))) &= \frac{\frac{z+1}{z-1}i+i}{\frac{z+1}{z-1}i-i} = \frac{\frac{z+1}{z-1}+1}{\frac{z+1}{z-1}-1} = \frac{(z+1)+(z-1)}{(z+1)-(z-1)} = \frac{2z}{2} = z. \end{align*}

From this, it follows that $z_{k+3} = z_k$, for all $k$. Thus $z_{2002} = z_{3\cdot 667+1} = z_1 = \frac{z_0+i}{z_0-i} = \frac{(\frac{1}{137}+i)+i}{(\frac{1}{137}+i)-i}= \frac{\frac{1}{137}+2i}{\frac{1}{137}}= 1+274i.$

Thus $a+b = 1+274 = \boxed{275}$.

See also

2002 AIME I (Problems • Answer Key • Resources)
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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