cos45+icos135+…i^n*cos(45+90n)+…i^40*cos3645=?
Sum of First 4 terms
sum_(r=1)^(r=4)S_r
=cos45+icos135+i^2cos(45+90xx2)+i^3cos(45+90xx3)
=cos45-icos45+cos45-icos45
=1/sqrt2-i/sqrt2+1/sqrt2-i/sqrt2=sqrt2(1-i)
Sum of Second 4 terms
sum_(r=5)^(r=9)S_r
=i^4cos(45+90xx4)+i^5cos(45+90xx5)+i^6cos(45+90xx6)+i^7cos(45+90xx7)
=cos45-icos45+cos45-icos45
=1/sqrt2-i/sqrt2+1/sqrt2-i/sqrt2=sqrt2(1-i)
Considering this way we get 10 groups of 4terms having same sum.
So our sum
S=sum_(r=1)^(r=40)S_r+i^40*cos3645
=10xxsqrt2(1-i)+cos(45+90xx40)
=10sqrt2(1-i)+cos45
=10sqrt2(1-i)+1/sqrt2