Problem:
A particle is located on the coordinate plane at (5,0). Define a move for the particle as a counterclockwise rotation of Ο/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 β£pβ£+β£qβ£.
Solution:
Let the coordinate plane be the complex plane, and let zkβ be the complex number that represents the position of the particle after k moves. Multiplying a complex number by cisΞΈ corresponds to a rotation of ΞΈ about the origin, and adding 10 to a complex number corresponds to a horizontal translation of 10 units to the right. Thus z0β=5, and zk+1β=Οzkβ+10, where Ο=cis(Ο/4), for kβ₯0. Then
In particular, z150β=5Ο150+10(Ο149+Ο148+β―+1). Note that Ο8=1 and Οk+4=cis((k+4)Ο/4)=cis(kΟ/4+Ο)=βcis(kΟ/4)=βΟk. Applying the second equality repeatedly shows that z150β=5Ο150+10(Ο149+Ο148+β―+1)=5Ο6+10(βΟ(β1)+Ο3+Ο2+Ο+1)=5Ο6+10(Ο3+Ο2). The last expression equals
Thus (p,q)=(β52β,52β+5), and β£pβ£+β£qβ£=102β+5. The required value is therefore the greatest integer less than or equal to 10β 1.414+5=19.14, which is 19β.