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

Dec 27, 2015

${\left(2 i\right)}^{\frac{1}{2}} = 1 + i$
and
${\left(2 i\right)}^{\frac{1}{2}} = - 1 - i$

#### Explanation:

Consider ${\left(2 i\right)}^{\frac{1}{2}} = \sqrt{2 i} = x + y i$ where $x$ and $y$ are real unknown numbers that we have to find.
Then $2 i = {\left(x + y i\right)}^{2} = {x}^{2} + 2 x y i + {\left(y i\right)}^{2} = \left({x}^{2} - {y}^{2}\right) + 2 x y i$
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
$0 = {x}^{2} - {y}^{2}$
$2 = 2 x y$

or, simplifying,
${x}^{2} = {y}^{2}$
$x y = 1$

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:
$\sqrt{2 i} = 1 + i$ or $\sqrt{2 i} = - 1 - i$
Check:
${\left(1 + i\right)}^{2} = 1 + 2 i + {i}^{2} = 1 + 2 i - 1 = 2 i$ (GOOD)
${\left(- 1 - i\right)}^{2} = {\left(- 1\right)}^{2} + 2 i + {\left(- i\right)}^{2} = 1 + 2 i - 1 = 2 i$ (GOOD)

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