Question: Find all values of (1+i)^(1/4) (its raise to 1/4)
De Moivre's Thoerem
(cosx + i.sinx)^n = cos(nx) + i.sin(nx)
For n an integer and x is any real or complex number.
De Moivre's theorem does not, in gerneral, hold for non-integer values, e.g. n = 1/4.
When n is a non-integer, the results for the rhs are multi-valued.
cos(π/4) = 1/√2 and so also does cos(2nπ + π/4) = 1/√2.
sin(π/4) = 1/√2 and so also does sin(2nπ + π/4) = 1/√2.
Then,
cos(2nπ + π/4) + i.sin(2nπ + π/4) = 1/√2 + i.(1/√2) = (1/√2)(1 + i)
i.e. (1 + i) = √2(cos(2nπ + π/4) + i.sin(2nπ + π/4))
(1 + i)^(1/4) = 2^(1/8)(cos(2nπ + π/4) + i.sin(2nπ + π/4)^(1/4) = 2^(1/8)(cos(nπ/2 + π/16) + i.sin(nπ/2 + π/16))
The different possible values for (1+i)^(1/4) will cycle over 4 different values for n = 0 to 3.
n = 0: 2^(1/8)(cos(/16) + i.sin(/16))
n = 1: 2^(1/8)(cos(π/2 + π/16) + i.sin(π/2 + π/16)) = 2^(1/8)(sin(π/16) + i.cos(π/16))
n = 2: 2^(1/8)(cos(π + π/16) + i.sin(π + π/16)) = 2^(1/8)(-cos(π/16) - i.sin(π/16))
n = 3: 2^(1/8)(cos(3π/2 + π/16) + i.sin(3π/2 + π/16)) = 2^(1/8)(-sin(π/16) - i.cos(π/16))