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| 1. |
Find square root of ( -i) of complex Number |
| Answer» Let {tex}x + yi = \\sqrt { - i} {/tex}Squaring both sides, we get(x + yi)2 = -ix2 - y2 + 2xyi = -iEquating the real and imaginary partsx2 - 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}\xa0x2 + y2 = 1 . . . (ii) [Neglecting (-) sign as x2 + y2 > 0]Solving (i) and (ii), we get{tex}{x^2} = \\frac{1}{2}{/tex}\xa0and {tex}{y^2} = \\frac{1}{2}{/tex}{tex}\\therefore x = \\pm \\frac{1}{{\\sqrt 2 }}{/tex}\xa0and {tex}y = \\pm \\frac{1}{{\\sqrt 2 }}{/tex}Since the sign of xy is (-) then if{tex}x = \\frac{1}{{\\sqrt 2 }}{/tex},\xa0{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} | |