Question

Find the square root of the complex number - 7-24i.​

Answer 1

Square root of complex numbers -

$+ - \sqrt{ \frac{ |z| + re(z)}{2} } - i\sqrt{ \frac{ |z| - re(z)}{2} }$

-7-24i = square root .

$|z| = \sqrt{ {re(z)}^{2} + {im(z)}^{2} }$

$|z| = \sqrt{7^{2} + {24}^{2} }$

|z| = 25 .

Now putting value in above mentioned formula .

$\sqrt{ \frac{25 + 7}{2} } - i \sqrt{ \frac{25 - 24}{2} }$

$\sqrt{16} - i \sqrt{ \frac{1}{2} }$

$4 - i \frac{1}{ \sqrt{2} }$

hence square root of-7-24i is

(+/-)(4-i(1/√2)

hope it helps.

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 ❣️