Problem:
For how many positive integers n less than or equal to 1000 is
(sint+icost)n=sinnt+icosnt
true for all real t?
Solution:
Note that
(sint+icost)n=[cos(2π−t)+isin(2π−t)]n=cosn(2π−t)+isinn(2π−t)=cos(2nπ−nt)+isin(2nπ−nt)
and that sinnt+icosnt=cos(2π−nt)+isin(2π−nt). Thus the given condition is equivalent to
cos(2nπ−nt)=cos(2π−nt) and sin(2nπ−nt)=sin(2π−nt)
In general, cosα=cosβ and sinα=sinβ if and only if α−β=2πk. Thus
2nπ−nt−2π+nt=2πk
which yields n=4k+1. Because 1≤n≤1000, conclude that 0≤k≤249, so there are 250 values of n that satisfy the given conditions.
OR
Observe that
(sint+icost)n=[i(cost−isint)]n=in(cosnt−isinnt), and that sinnt+icosnt=i(cosnt−isinnt)
Thus the given equation is equivalent to in(cosnt−isinnt)=i(cosnt−isinnt). This is true for all real t when in=i. Thus n must be 1 more than a multiple of 4, so there are 250 values of n that satisfy the given conditions.
The problems on this page are the property of the MAA's American Mathematics Competitions