Problem:
Let F(z)=zβiz+iβ for all complex numbers zξ =i, and let znβ=F(znβ1β) for all positive integers n. Given that z0β=1371β+i and z2002β=a+bi, where a and b are real numbers, find a+b.
Solution:
Calculate
F(F(z))=zβiz+iββizβiz+iβ+iβ=z+iβizβ1z+i+iz+1β=1βi1+iββ
zβ1z+1β=iβ
zβ1z+1β
and
F(F(F(z)))=F(iβ
zβ1z+1β)=iβ
zβ1z+1ββiiβ
zβ1z+1β+iβ=z+1β(zβ1)z+1+zβ1β=z
which shows that znβ=znβ3β for all nβ₯3. In particular, z2002β=z2002β667β
3β= z1β=1371β1371β+2iβ=1+1371β2βi=1+274i, and a+b=275β.
The problems on this page are the property of the MAA's American Mathematics Competitions