Question

square root of a complex number 5-12i​

Answer 1

Answer :

± (3 - 2i)

Solution :

• Given complex no. : 5 - 12i

• To find : √(5 - 12i)

Here ,

The given complex no. is (5 - 12i) .

We need to find the square root of the given complex no. (5 - 12i) .

Let the square root of the given complex no. be (x - iy) .

Thus ,

=> x - iy = √(5 - 12i)

=> (x - iy)² = 5 - 12i

=> x² + (iy)² - 2•x•iy = 5 - 12i

=> x² + i²y² - 2xyi = 5 - 12i

=> x² - y² - 2xyi = 5 - 12i

Now ,

Comparing the real and imaginary parts both the sides of about equation ,

We have ;

• x² - y² = 5 -------(1)

• 2xy = 12

→ y = 12/2x

→ y = 6/x ---------(2)

Now ,

Putting y = 6/x in eq-(1) , we have ;

=> x² - y² = 5

=> x² - (6/x)² = 5

=> x² - 36/x² = 5

=> (x⁴ - 36)/x² = 5

=> x⁴ - 36 = 5x²

=> x⁴ - 5x² - 36 = 0

=> x⁴ - 9x² + 4x² - 36 = 0

=> x²(x² - 9) + 4(x² - 9) = 0

=> (x² - 9)(x² + 4) = 0

=> x² - 9 = 0 { °•° x² + 4 ≠ 0 }

=> x² = 9

=> x² = √9

=> x = ±3

Now ,

Putting x = ±3 in eq-(2) , we have ;

=> y = 6/x

=> y = 6/±3

=> y = ±2

Thus ,

√(5 - 12i) = ± (3 - 2i)

Hence ,

Required answer is : ± (3 - 2i)

Answer 2

$let \sqrt{ - 5 + 12i} = a + ib$

$- 5 + 12i = (a + ib {)}^{2} = {a}^{2} - {b}^{2} + 2iab$

$We \: will \: get \: {a}^{2} - {b}^{2} = - 5, \: 2ab = 12$

${a}^{2} - \frac{36}{ {a}^{2} } = - 5.......b = \frac{6}{a}$

${a}^{4} + 5 {a}^{2} - 36 = 0$

$( {a}^{2} - 4)( {a}^{2} + 9) = 0$

$a = ±2 \: and \: ±3$

$we \: have \: a = 2 \: and \: b = 3$

$a = - 2 \: and \: b = - 3$

$b = \frac{6}{a} = ±3 so \sqrt{ - 5 + 12i}$

$= ±(2 + 3i)$

hope it's helps you ❤️