Problem:
A sequence of complex numbers z0,z1,z2,… is defined by the rule
zn+1=znizn
where zn is the complex conjugate of zn and i2=−1. Suppose that ∣z0∣=1 and z2005=1. How many possible values are there for z0 ?
Answer Choices:
A. 1
B. 2
C. 4
D. 2005
E. 22005
Solution:
Note that
zn+1=znizn=znznizn2=∣zn∣2izn2
Since ∣z0∣=1, the sequence satisfies
z1=iz02,z2=iz12=i(iz02)2=−iz04
and, in general, when k≥2,
zk=−iz02k
Hence z0 satisfies the equation 1=−iz0(22005), so z0(22005)=i. Because every nonzero complex number has n distinct nth roots, this equation has 22005 solutions. So there are 22005 possible values for z0.
OR
Define
cisθ=cosθ+isinθ
Then if zn=r cis θ we have
zn+1=cis(−θ)cis(θ+90∘)=cis(2θ+90∘)
The first terms of the sequence are z0=cisα,z1=cis(2α+90∘)=iz02, z2=cis(4α+270∘)=cis(4α−90∘)=iz04,z3=cis(8α−90∘)=iz08, and, in general,
zn=iz0(2n) for n≥2
So
z2005=iz0(22005)=1 and z0(22005)=i
As before, there are 22005 possible solutions for z0.
The problems on this page are the property of the MAA's American Mathematics Competitions