Problem:
A particle is located on the coordinate plane at (5,0). Define a move for the particle as a counterclockwise rotation of π/4 radians about the origin followed by a translation of 10 units in the positive x-direction. Given that the particle's position after 150 moves is (p,q), find the greatest integer less than or equal to ∣p∣+∣q∣.
Solution:
Let the coordinate plane be the complex plane, and let zk be the complex number that represents the position of the particle after k moves. Multiplying a complex number by cisθ corresponds to a rotation of θ about the origin, and adding 10 to a complex number corresponds to a horizontal translation of 10 units to the right. Thus z0=5, and zk+1=ωzk+10, where ω=cis(π/4), for k≥0. Then
In particular, z150=5ω150+10(ω149+ω148+⋯+1). Note that ω8=1 and ωk+4=cis((k+4)π/4)=cis(kπ/4+π)=−cis(kπ/4)=−ωk. Applying the second equality repeatedly shows that z150=5ω150+10(ω149+ω148+⋯+1)=5ω6+10(−ω(−1)+ω3+ω2+ω+1)=5ω6+10(ω3+ω2). The last expression equals
Thus (p,q)=(−52,52+5), and ∣p∣+∣q∣=102+5. The required value is therefore the greatest integer less than or equal to 10⋅1.414+5=19.14, which is 19.