Problem:
For any complex number w=a+bi,∣w∣ is defined to be the real number a2+b2. If w=cos40∘+isin40∘, then ∣∣∣w+2w2+3w3+⋯+9w9∣∣∣−1 equals
Answer Choices:
A. 91sin40∘
B. 92sin20∘
C. 91cos40∘
D. 181cos20∘
E. none of these
Solution:
We derive a more general result: if n is an integer >1 and w=cosn2π+isinn2π, then
∣∣∣w+2w2+⋯+nwn∣∣∣−1=n2sinnπ.
To prove this, let S=w+2w2+⋯+nwn. Multiplying both sides of this equation by w and subtracting Sw from S, we get
S(1−w)=w+w2+⋯+wn−nwn+1.
Now, w=1 since n>1. Thus we may use the formula for summing a geometric series to obtain
S(1−w)=w−1wn+1−w−nwn+1.
Since wn=1 (by De Moivre's formula), this reduces further to S(1−w)=−nw. Thus
S1=nww−1,∣S∣1=n∣w−1∣.
Finally, ∣w−1∣ is the length of the side of the regular n-gon inscribed in the unit circle, since 1 and w are consecutive vertices. It is well known that this side length is 2sinnπ (in the isosceles triangle with vertices 0,1,w, drop an altitude from 0).