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2002 AIME I Problems/Problem 12

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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)}{(...

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...

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

See also

2002 AIME I (ProblemsResources)
Preceded by
Problem 11
Followed by
Problem 13
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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