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

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== See also ==
 
== See also ==
 
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Revision as of 18:55, 4 July 2013

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:

$F(z)=z+iziF(F(z))=z+izi+iz+izii=(z+i)+i(zi)(z+i)i(zi)=z+i+zi+1z+izi1=(z+1)(i+1)(z1)(1i)=(z+1)(i+1)2(z1)(12+12)=(z+1)(2i)(z1)(2)=z+1z1iF(F(F(z)))=z+1z1i+iz+1z1ii=z+1z1+1z+1z11=(z+1)+(z1)(z+1)(z1)=2z2=z.$ (Error compiling LaTeX. Unknown error_msg)

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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