Question: use demoviers therom prove (1+i)^8 +(1-i)^8 =32.
DeMoivre's Theorem
(cosx + i.sinx)^n = cos(nx) + i.sin(nx)
where n is an integer and x is any real or complex number.
cos(π/4) = sin(π/4) = 1/√2.
So, cos(π/4) + i.sin(π/4) = (1/√2)(1 + i)
Therefore, (1 + i) = √2(cos(π/4) + i.sin(π/4) )
(1 + i)^8 = (2^4)(cos(π/4) + i.sin(π/4) )^8 = 16(cos(2π) + i.sin(2π) ) = 16*1 = 16
Continuing,
cos(-π/4) = 1/√2. sin(-π/4) = -1/√2.
So, cos(-π/4) + i.sin(-π/4) = (1/√2)(1 - i)
i.e. cos(π/4) - i.sin(π/4) = (1/√2)(1 - i)
Therefore, (1 - i) = √2(cos(π/4) - i.sin(π/4) )
(1 - i)^8 = (2^4)(cos(π/4) - i.sin(π/4) )^8 = 16(cos(2π) - i.sin(2π) ) = 16*1 = 16
Finally,
(1+i)^8 + (1-i)^8 = 16 + 16 = 32