Let \(F(z)=\frac{z+i}{z-i}\) for all complex numbers \(z\not= i\), and let \(z_n=F(z_{n-1})\) for all positive integers \(n\). Given that \(z_0=\frac 1{137}+i\) and \(z_{2002}=a+bi\), where \(a\) and \(b\) are real numbers, find \(a+b\).

(第二十届AIME1 2002 第12题)