Express(1+i)^20 in the form of a+ib
we could expand this using the coefficients from Pascal triangle, but that
would be quite tedious.
let's use De Moivre's Theorem:
let z = 1 + i
argument = √(1+1) = √2
angle : tan^-1 (1/1) = π /4 or 45°
z = √2(cos π/4 + i sin π/4)
z^20 = (√2)^20 (cos 5π + isin 5π) = 1024(-1 + 0) = -1024 + 0i