Problem:
Let F(z)=z−iz+i​ for all complex numbers zî€ =i, and let zn​=F(zn−1​) for all positive integers n. Given that z0​=1371​+i and z2002​=a+bi, where a and b are real numbers, find a+b.
Solution:
Calculate
F(F(z))=z−iz+i​−iz−iz+i​+i​=z+i−iz−1z+i+iz+1​=1−i1+i​⋅z−1z+1​=i⋅z−1z+1​
and
F(F(F(z)))=F(i⋅z−1z+1​)=i⋅z−1z+1​−ii⋅z−1z+1​+i​=z+1−(z−1)z+1+z−1​=z
which shows that zn​=zn−3​ for all n≥3. In particular, z2002​=z2002−667⋅3​= z1​=1371​1371​+2i​=1+1371​2​i=1+274i, and a+b=275​.
The problems on this page are the property of the MAA's American Mathematics Competitions