What is the value of *sigma sum sign* n=0 to 40 (i^n*cos (45 + 90n)°?
39
∑ i^n cos(45+90n)°
n=0
= 1 cos45° + i cos135° - 1 cos225° - i cos 315° + ...
= cos45° - i cos45° + cos45° - i cos45°
... repeated for 10 cycles
= 0
so the sum for n = 1..40 is just cos45° = 1/√2