Find square root of ( -i) …

Let {tex}x + yi = \sqrt { - i} {/tex}
Squaring both sides, we get
(x + yi)² = -i
x² - y² + 2xyi = -i
Equating the real and imaginary parts
x² - y² = 0 . . . (i)
and 2xy = -1
{tex}\therefore xy = - \frac{1}{2}{/tex}
Now using the identity
(x² + y²)² = (x² - y²)² + 4x²y²
{tex} = {(0)^2} + 4{\left( { - \frac{1}{2}} \right)^2}{/tex}
= 1
{tex}\therefore {/tex} x² + y² = 1 . . . (ii) [Neglecting (-) sign as x² + y² > 0]
Solving (i) and (ii), we get
{tex}{x^2} = \frac{1}{2}{/tex} and {tex}{y^2} = \frac{1}{2}{/tex}
{tex}\therefore x = \pm \frac{1}{{\sqrt 2 }}{/tex} and {tex}y = \pm \frac{1}{{\sqrt 2 }}{/tex}
Since the sign of xy is (-) then if
{tex}x = \frac{1}{{\sqrt 2 }}{/tex}, {tex}y = - \frac{1}{{\sqrt 2 }}{/tex}
If {tex}x = - \frac{1}{{\sqrt 2 }}\;y = \frac{1}{{\sqrt 2 }}{/tex}
{tex}\style{font-family:Tahoma}{\style{font-size:8px}{\therefore\sqrt{-i}=\pm\left(\frac1{\sqrt2}-\frac1{\sqrt2}i\right)}}{/tex}