AIME 2002 I · 第 12 题

Solution

Iterating $F$ we get:

$$\begin{aligned} F(z) &= \frac{z+i}{z-i}\\ F(F(z)) &= \frac{\frac{z+i}{z-i}+i}{\frac{z+i}{z-i}-i} = \frac{(z+i)+i(z-i)}{(z+i)-i(z-i)}= \frac{z+i+zi+1}{z+i-zi-1}= \frac{(z+1)(i+1)}{(z-1)(1-i)}\\ &= \frac{(z+1)(i+1)^2}{(z-1)(1^2+1^2)}= \frac{(z+1)(2i)}{(z-1)(2)}= \frac{z+1}{z-1}i\\ F(F(F(z))) &= \frac{\frac{z+1}{z-1}i+i}{\frac{z+1}{z-1}i-i} = \frac{\frac{z+1}{z-1}+1}{\frac{z+1}{z-1}-1} = \frac{(z+1)+(z-1)}{(z+1)-(z-1)} = \frac{2z}{2} = z. \end{aligned}$$ 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 $a+b = 1+274 = \boxed{275}$.

Solution 2 (Very Risky, only attempt if you have very little time left)

Solving for $z_{1}$, we have $\frac{\frac{1}{137}+2i}{\frac{1}{137}}=1+274i$. Notice that when we solve for $z_{2}$, the result is equal to some non-integral real and imaginary part, where the value of $a+b$ for $z_{2}$ is not equal to an integer. Because this is an AIME problem, the answer must be an integer, so there is most likely a pattern or cycle that $z_{n}$ goes through, and because out of the first 3 terms, only one of which has an integral value for $a+b$, you can guess that the answer is likely $1+274=\boxed{275}$

~Voidling