Question

Q27) Find the square root of (7+24i)

Answer 1

hey dear

here is your answer

Solution

√7+ 24i

Let x+ iy = √ 7+ 24i

squaring on both the sides

( x+ iy) ^2 = 7+ 24i. ( therefore i^2 = -1)

x^2 - y^2 + I. 2xy = 7+ 24i

comparing real and imaginary part

x^2 - y^2 = 7

2xy = 24

xy = 12

Also

( x^2 + y^2)^2 = ( x^2 - y^2)^2 + (2xy) ^2

( x^2 + y^2)^2 = 49 + 576 = 625

x^2 + y^2 = 25. ( since square of two no not - )

so

x^2 + y^2 = 25
x^2 - y^2 = 7

(adding both the equation we get )

2x^2 = 32

x^2 = 16

x = 4

When x = 4

and y = 12/4

y = 3

When x = -4 , y = -3

So √7+ 24i

= 4+3i

and - 4 -3i

both are the positive negative answer


hope it helps

thank you

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 ❣️