Problem:
For any complex number w=a+bi,β£wβ£ is defined to be the real number a2+b2β. If w=cos40β+isin40β, then β£β£β£βw+2w2+3w3+β―+9w9β£β£β£ββ1 equals
Answer Choices:
A. 91βsin40β
B. 92βsin20β
C. 91βcos40β
D. 181βcos20β
E. none of these
Solution:
We derive a more general result: if n is an integer >1 and w=cosn2Οβ+isinn2Οβ, then
To prove this, let S=w+2w2+β―+nwn. Multiplying both sides of this equation by w and subtracting Sw from S, we get
S(1βw)=w+w2+β―+wnβnwn+1.
Now, wξ =1 since n>1. Thus we may use the formula for summing a geometric series to obtain
S(1βw)=wβ1wn+1βwββnwn+1.
Since wn=1 (by De Moivre's formula), this reduces further to S(1βw)=βnw. Thus
S1β=nwwβ1β,β£Sβ£1β=nβ£wβ1β£β.
Finally, β£wβ1β£ 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 2sinnΟβ (in the isosceles triangle with vertices 0,1,w, drop an altitude from 0).