Find the square roots of the following:1. 1-i2. -i3. i4. 1+i

1.	Find the square roots of the following:1. 1-i2. -i3. i4. 1+i
Answer» 1. Let {tex}x + yi = \\sqrt {1 - i} {/tex}Squaring both sides, we getx2 - y2 + 2xyi = 1 - iEquating the real and imaginary partsx2 - y2 = 1 and 2xy=-1...\xa0. (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}\xa0. . . . (ii) [Neglecting (-) sign as x2 + y2 > 0]Solving (i) and (ii) we get{tex}{x^2} = \\frac{{\\sqrt 2 + 1}}{2}{/tex}\xa0and {tex}y = \\frac{{\\sqrt 2 - 1}}{2}{/tex}{tex}\\therefore x = \\pm \\sqrt {\\frac{{\\sqrt 2 + 1}}{2}} {/tex}\xa0and {tex}y = \\pm \\sqrt {\\frac{{\\sqrt 2 - 1}}{2}} {/tex}Since the sign of xy is negative.{tex}\\therefore {/tex}\xa0if {tex}x = \\sqrt {\\frac{{\\sqrt 2 + 1}}{2}} {/tex}\xa0then {tex}y = - \\sqrt {\\frac{{\\sqrt 2 - 1}}{2}} {/tex}and if {tex}x = - \\sqrt {\\frac{{\\sqrt 2 + 1}}{2}} {/tex}\xa0then {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}\xa0

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