Problem:
A sequence of complex numbers z0β,z1β,z2β,β¦ is defined by the rule
zn+1β=znββiznββ
where znββ is the complex conjugate of znβ and i2=β1. Suppose that β£z0ββ£=1 and z2005β=1. How many possible values are there for z0β ?
Answer Choices:
A. 1
B. 2
C. 4
D. 2005
E. 22005
Solution:
Note that
zn+1β=znββiznββ=znβznββizn2ββ=β£znββ£2izn2ββ
Since β£z0ββ£=1, the sequence satisfies
z1β=iz02β,z2β=iz12β=i(iz02β)2=βiz04β
and, in general, when kβ₯2,
zkβ=βiz02kβ
Hence z0β satisfies the equation 1=βiz0(22005)β, so z0(22005)β=i. Because every nonzero complex number has n distinct nth roots, this equation has 22005 solutions. So there are 22005 possible values for z0β.
OR
Define
cisΞΈ=cosΞΈ+isinΞΈ
Then if znβ=r cis ΞΈ we have
zn+1β=cis(βΞΈ)cis(ΞΈ+90β)β=cis(2ΞΈ+90β)
The first terms of the sequence are z0β=cisΞ±,z1β=cis(2Ξ±+90β)=iz02β, z2β=cis(4Ξ±+270β)=cis(4Ξ±β90β)=iz04ββ,z3β=cis(8Ξ±β90β)=iz08ββ, and, in general,
znβ=iz0(2n)ββ for nβ₯2
So
z2005β=iz0(22005)ββ=1 and z0(22005)β=i
As before, there are 22005β possible solutions for z0β.
The problems on this page are the property of the MAA's American Mathematics Competitions