Question

find the square root o the complex number(7-24i)

Answer 1

let x+iy=underroot 7-24i, then(x+iy) ka whole square=7-24i, or x square -y square +2xyi=7-24i, equating real and imaginary parts, we have x square -y square=7 eq.(1), 2xy=-24, we know the identity (x square + y square)ka whole square=(x square -y square) ka whole square +(2xy) ka whole square=49+576=625, thus,x square +y square =25 eq(2), from (1) and (2), x square=16 and y square=9 or x=+ or - (4) and y=+or-(3), since the product xy is negative, we have x=4, y=-3 or,x=-4,y=3, thus, the square roots of 7-24i are 4-3i and -4+3i please mark it as a brainiest answer.

Answer 2

ANSWER:-

$let \sqrt{ - 7 - 24i} = a + ib$

$- 7 - 24i = (a + ib {)}^{2} = {a}^{2} - {b}^{2} + 2iab$

$comparing \: coeffiecient \: we \: get$

${a}^{2} - {b}^{2} = - 7 \: \: and \: \: 2ab = - 24$

$ab = - 12$

$b = \frac{ - 12}{a}$

${a}^{2} - \frac{144}{ {a}^{2} } = - 7$

${a}^{2} + 7 {a}^{2} - 144 = 0$

$= > ( {a}^{2} - 9)( {a}^{2} + 16) = 0$

$Hence, {a}^{2} + 16≠0 \: \: \: so, {a}^{2} = 9$

$a = ±3$

$a = \frac{ - 12}{a} = ±4$

$for \: a = 3,b = - 4$

$a = - 3,b = - 4$

$so, = \sqrt{ - 7 - 24i} = ±(3 - 4i)$

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