Question

square roots of the complex numbers 7+24i​

Answer 1

Answer:

lemme solve it in an easier way

here you go

$\sqrt{7 + 24i} = \sqrt{16 - 9 + 24i} (breaking \: 7in \: terms \\ \: of \: difference \: of \: two \: squares) \\ = \sqrt{ {4}^{2} + {3i}^{2} + 2.4.3i} \\ = \sqrt{ {(4 + 3i)}^{2} } = 4 + 3i$

Answer 2

ANSWER:-

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

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

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

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

$ab = - 12$

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

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

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

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

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

$a = ±3$

$a = \frac{ - 12}{a} = ±4$

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

$a = - 3,b = - 4$

$so, = \sqrt{ - 7 - 24i} = ±(3 - 4i)$

HOPE IT'S HELPS YOU ❣️