How do you simplify (2i)^(1/2)(2i)12?

1 Answer
Zor Shekhtman
Dec 27, 2015

There are two answers:
(2i)^(1/2) = 1+i(2i)12=1+i
and
(2i)^(1/2) = -1-i(2i)12=−1−i

Explanation:

Consider (2i)^(1/2) = sqrt(2i) = x+yi(2i)12=√2i=x+yi where xx and yy are real unknown numbers that we have to find.
Then 2i = (x+yi)^2 = x^2+2xyi+(yi)^2 = (x^2-y^2)+2xyi2i=(x+yi)2=x2+2xyi+(yi)2=(x2−y2)+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^20=x2−y2
2 = 2xy2=2xy

or, simplifying,
x^2 = y^2x2=y2
xy = 1xy=1

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

  1. If x=yx=y, from the second equation follows that
    x^2=1x2=1 and either x=1x=1 or x=-1x=−1
    So, we have two solutions:
    sqrt(2i) = 1+i√2i=1+i or sqrt(2i) = -1-i√2i=−1−i
    Check:
    (1+i)^2 = 1+2i+i^2 = 1+2i-1 = 2i(1+i)2=1+2i+i2=1+2i−1=2i (GOOD)
    (-1-i)^2 = (-1)^2+2i+(-i)^2 = 1+2i-1 = 2i(−1−i)2=(−1)2+2i+(−i)2=1+2i−1=2i (GOOD)

  2. If x=-yx=−y, from the second equation follows that
    -x^2=1−x2=1, which has no solutions among real numbers.

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