Question

find the square root of
1-i​

Answer 1

${\bold {\huge {Question-}}}$

$\sqrt{1 - i}$

${\bold {\large{\green {\underline{Answer:}}}}}$

let us assume (x+iy) be the roots of 1-i

now,

$1 - i \: = {(x + iy)}^{2} \\ 1 - i = {x}^{2} - {y}^{2} +( 2xy)i$

on comparing:

${x }^{2} - {y}^{2} = 1 - - - - - (1) \\ 2xy = - 1 - - - - - - (2)$

we know that

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

by equation[ 1 ]and[ 2 ]

$( {x}^{2} + {y}^{2} )^{2} = 1 + 1$

${x}^{2} + {y}^{2} = \sqrt{2} - - - - - (3)$

on adding equ. 1 and 3

${x}^{2} = \frac{1 + \sqrt{2} }{2}$

$x = + - \sqrt{ \frac{2 + \sqrt{2} }{2} }$

$y = + - \sqrt{ \frac{2 - \sqrt{2} }{2} }$

so the roots will be:

$\sqrt{ \frac{2 + \sqrt{2} }{2} } - i \sqrt{ \frac{2 - \sqrt{2} }{2} }$

or

$\sqrt{ \frac{2 + \sqrt{2} }{2} } + i \sqrt{ \frac{2 - \sqrt{2} }{2} }$