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Sia 🤖 1 month, 2 weeks ago

1. Let {tex}x + yi = \sqrt {1 - i} {/tex}

Squaring both sides, we get

x

^{2}- y^{2}+ 2xyi = 1 - iEquating the real and imaginary parts

x

^{2}- y^{2}= 1 and 2xy=-1... . (i){tex}\therefore \;xy = \frac{{ - 1}}{2}{/tex}

Using the identity

(x

^{2}+ y^{2})^{2}= (x^{2}- y^{2})^{2}+ 4x^{2}y^{2}{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 x

^{2}+ y^{2}> 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}

0Thank You