Find the square roots of the …

1. Let {tex}x + yi = \sqrt {1 - i} {/tex}
Squaring both sides, we get
x² - y² + 2xyi = 1 - i
Equating the real and imaginary parts
x² - y² = 1 and 2xy=-1... . (i)
{tex}\therefore \;xy = \frac{{ - 1}}{2}{/tex}
Using the identity
(x² + y²)² = (x² - y²)² + 4x²y²
{tex} = {\left( 1 \right)^2} + 44{\left( { - \frac{1}{2}} \right)^2}{/tex}
= 1 + 1
= 2
{tex}\therefore \;{x^2} + {y^2} = \sqrt 2 {/tex} . . . . (ii) [Neglecting (-) sign as x² + y² > 0]
Solving (i) and (ii) we get
{tex}{x^2} = \frac{{\sqrt 2 + 1}}{2}{/tex} and {tex}y = \frac{{\sqrt 2 - 1}}{2}{/tex}
{tex}\therefore x = \pm \sqrt {\frac{{\sqrt 2 + 1}}{2}} {/tex} and {tex}y = \pm \sqrt {\frac{{\sqrt 2 - 1}}{2}} {/tex}
Since the sign of xy is negative.
{tex}\therefore {/tex} if {tex}x = \sqrt {\frac{{\sqrt 2 + 1}}{2}} {/tex} then {tex}y = - \sqrt {\frac{{\sqrt 2 - 1}}{2}} {/tex}
and if {tex}x = - \sqrt {\frac{{\sqrt 2 + 1}}{2}} {/tex} then {tex}y = \sqrt {\frac{{\sqrt 2 - 1}}{2}} {/tex}
{tex}\therefore \sqrt {1 - i} = \pm \left( {\sqrt {\frac{{\sqrt 2 + 1}}{2}} - \sqrt {\frac{{\sqrt 2 - 1}}{2}} i} \right){/tex}