Question

square root of 2-2i✓3

Answer 1

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Answer 2

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$let \: \sqrt{(2 - 2i \sqrt{3}) } = a + ib$
$(2 - 2i \sqrt{3} ) = {(a + ib)}^{2}$
$2 - 2i \sqrt{3 } = {a}^{2} - {b}^{2} + i(2ab)$
by equating real to real and imaginary to imaginary part we get,

$2 = {a}^{2} - {b}^{2} \: \: \: \: \: \: \: - (i)$
$- 2 \sqrt{3} = 2ab$
$b = - \frac{ \sqrt{3} }{a}$
now, in eq. (i)
$2 = {a}^{2} - {( - \frac{ \sqrt{3} }{a}) }^{2}$

$2 {a}^{2} = {a}^{4} - 3$
${a}^{4} - 2 {a}^{2} - 3 = 0$
${a}^{4} - 3 {a}^{2} + {a}^{2} - 3 = 0$
${a}^{2} ( {a}^{2} - 3) + 1( {a}^{2} - 3) = 0$
${a}^{2} = 3 \: \: or \: \: {a}^{2} = - 1$
we can't take imaginary value of 'a'
so,

$({a}^{2} = - 1 )\: is \: not \: possible \:$
$a = (\binom{ + }{ - } \sqrt{3} )$

so
$b = ( \binom{ - }{ + } 1)$
$\sqrt{(2 - 2i \sqrt{3}) } = \binom{ + }{ - } ( \sqrt{3} - 1i)$
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