2002 AIME I Problems/Problem 12

Revision as of 22:20, 4 May 2008 by Jam (talk | contribs) (solution)

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:

$\begin{align*} F(z) &= \frac{z+i}{z-i}\\ F(F(z)) &= \frac{\frac{z+i}{z-i}+i}{\frac{z+i}{z-i}-i} = \frac{(z+i)+i(z-i)}{(z+i)-i(z-i)}= \frac{z+i+zi+1}{z+i-zi-1}= \frac{(z+1)(i+1)}{(z-1)(1-i)}\\ &= \frac{(z+1)(i+1)^2}{(z-1)(1^2+1^2)}= \frac{(z+1)(2i)}{(z-1)(2)}= \frac{z+1}{z-1}i\\ F(F(F(z))) &= \frac{\frac{z+1}{z-1}i+i}{\frac{z+1}{z-1}i-i} = \frac{\frac{z+1}{z-1}+1}{\frac{z+1}{z-1}-1} = \frac{(z+1)+(z-1)}{(z+1)-(z-1)} = \frac{2z}{2} = z. \end{align*}$ (Error compiling LaTeX. ! Package amsmath Error: \begin{align*} allowed only in paragraph mode.)

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 (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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