if a=cos 2pi/5 + i sin2pi/5, find a+a2+a3+a4 Share with your friends Share 2 Tanveer Sofi answered this We havea=cos2π5+i sin2π5Now using De Moivre's theorem if n is a positive integer, we havean=cos2π5+i sin2π5n=cos2nπ5+i sin2nπ5Therefore,we havea+a2+a3+a4=cos2π5+i sin2π5+cos2π5+i sin2π52+cos2π5+i sin2π53+cos2π5+i sin2π54=cos2π5+i sin2π5+cos2×2π5+i sin2×2π5+cos3×2π5+i sin3×2π5+cos4×2π5+i sin4×2π5=cos2π5+cos4π5+cos6π5+cos8π5+i sin2π5+sin4π5+sin6π5+sin8π5Now, using cosC+cosD=2cosC+D2 cosC-D2 and sinC+sinD=2sinC+D2 cosC-D2, we havecos2π5+cos4π5+cos6π5+cos8π5+i sin2π5+sin4π5+sin6π5+sin8π5=2cos3π5 cosπ5+2cos7π5 cosπ5+i2sin3π5 cosπ5+2sin7π5 cosπ5=2cosπ5cos3π5 +cos7π5+isin3π5 +sin7π5=2cosπ52cos7π5+3π52 cos7π5-3π52+i2sin7π5+3π52 +sin7π5-3π52=2cosπ52cosπ cos2π5+i2sinπ sin2π5=-4cosπ5 cos2π5 As, cosπ=-1 and sinπ=0 -1 View Full Answer
