Solution:

Composing the map with itself, $$F(F(z)) = \frac{\frac{z+i}{z-i} + i}{\frac{z+i}{z-i} - i} = \frac{(z + i) + i(z - i)}{(z + i) - i(z - i)} = \frac{(1 + i)(z + 1)}{(1 - i)(z - 1)} = i\,\frac{z + 1}{z - 1},$$ and applying $F$ once more gives $$F(F(F(z))) = \frac{i\,\frac{z+1}{z-1} + i}{i\,\frac{z+1}{z-1} - i} = \frac{(z + 1) + (z - 1)}{(z + 1) - (z - 1)} = z,$$ so the sequence $z_0, z_1, z_2, \ldots$ is periodic with period $3.$

Since $2002 = 3 \cdot 667 + 1,$ we have $z_{2002} = z_1 = F(z_0) = \frac{z_0 + i}{z_0 - i} = \frac{\frac{1}{137} + 2i}{\frac{1}{137}} = 1 + 274i.$ Thus $a + b = 1 + 274 = 275.$