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

1 Answer
Dec 27, 2015

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.