Complex numbers?

Question

Find the two values of (-i)^1/2 in x+iy form where x and y are real numbers

FInd four values of (-i)^1/4 in x+iy form

Answer 1

You seek a number x + iy such that (x+iy)^2 = i = (0 + i1)

(x + iy)^2 = x^2 + 2xiy + (iy)^2 =
x^2 + 2xiy - y^2

You want
x^2 - y^2 = 0
and
2xy = 1

Two unknowns, two equations, go for it.

Answer 2

a)
0.70711-0.70711i
-0.707110.70711i


b)
0.92388-0.38268i
-0.923880.38268i
0.382680.92388i
-0.38268-0.92388i

How to solve it:

Just work in polars:
-i = 1 (-90) or 1 (270)
so -i ^1/2 = 1(-45) and 1 (135)

and, the same way:
-i^1/4 = (-i^1/2)^-1/2 = 1 (-22.5), 1 (157.5), 1 (67.5) and 1 (-112.5)

remember that 1 (-45) = 1 (315); and that 1 (135) = 1 (-225)

Answer 3

http://en.wikipedia.org/wiki/De_Moivre%27s_formula