De Moivre's theorem is helpful but, to use this theorem, you need to convert the algebraic from of complex number 1 + i into its polar form, using the following formulas, such that:

`z = x + i*y`

`z = sqrt(x^2+y^2)(cos(tan^(-1)(y/x)) + i*sin (tan^(-1)(y/x)))`

Identifying `x = 1` and `y = 1` and reasoning by analogy, yields:

`z = sqrt(1^2+1^2)(cos (tan^(-1)(1)) + i*sin (tan^(-1)(1)))`

`z = sqrt(2)(cos (pi/4) + i*sin (pi/4))`

Raising to the given power yields:

`z^(1/4) = (sqrt(2))^(1/4)(cos (pi/4) + i*sin (pi/4))^(1/4)`

Using De Moivre's theorem yields:

`z^(1/4) = (root(8)(2))(cos ((pi/4 + 2npi)/4) + i*sin ((pi/4 + 2npi)/4))`

`z^(1/4) = (root(8)(2))(cos (pi/16 + npi/2) + i*sin (pi/16 + npi/2)), k = 0,1,2,3`

**Hence, evaluating `(1 + i)^(1/4)` yields **`z^(1/4) = (root(8)(2))(cos (pi/16 + npi/2) + i*sin (pi/16 + npi/2)), k = 0,1,2,3.`