Question

How do you simplify `(2i)^(1/2)`?

Answer 1

There are two answers:
`(2i)^(1/2) = 1+i`
and
`(2i)^(1/2) = -1-i`

Explanation:

Consider `(2i)^(1/2) = sqrt(2i) = x+yi` where `x` and `y` are real unknown numbers that we have to find.
Then `2i = (x+yi)^2 = x^2+2xyi+(yi)^2 = (x^2-y^2)+2xyi`
Therefore, equating real and imaginary parts separately for left and right sides of this equation, we get a system of two equations with two unknowns
`0 = x^2-y^2`
`2 = 2xy`

or, simplifying,
`x^2 = y^2`
`xy = 1`

From the first equation we conclude that either `x=y` or `x=-y`.

1. If `x=y`, from the second equation follows that
   `x^2=1` and either `x=1` or `x=-1`
   So, we have two solutions:
   `sqrt(2i) = 1+i` or `sqrt(2i) = -1-i`
   Check:
   `(1+i)^2 = 1+2i+i^2 = 1+2i-1 = 2i` (GOOD)
   `(-1-i)^2 = (-1)^2+2i+(-i)^2 = 1+2i-1 = 2i` (GOOD)

2. If `x=-y`, from the second equation follows that
   `-x^2=1`, which has no solutions among real numbers.