Question

find the square root of the complex number -8-6i​

Answer 1

Answer:

Step by step Explanation:

The square root of the given complex number can be expressed as a complex number x + iy.

We have x + yi = sqrt (8 + 6i)

take the square of both the sides

=> (x + yi) ^2 = 8 + 6i

=> x^2 + y^2*i^2 + 2xyi = 8 + 6i

equate the real and complex coefficients

x^2 – y^2 = 8 and 2xy = 6

2xy = 6

=> xy = 3

=> x = 3/y

Substitute in x^2 – y^2 = 8

=> (3/y) ^2 – y^2 = 8

=> 9/y^2 – y^2 = 8

=> 9 – y^4 = 8y^2

=> y^4 + 8y^2 – 9 = 0

=> y^4 + 9y^2 – y^2 – 9 = 0

=> y^2(y^2 + 9) – 1(y^2 + 9) = 0

=> (y^2 – 1) (y^2 + 9) = 0

=> y^2 = 1 and y^2 = -9

we leave out y^2 = -9 as the coefficient y is real

y^2 = 1

=> y = 1, x = 3

and y = -1, x = -3

The square root of 8 + 6i = 3 + i ; -3 – i

Answer 2

$\Large\bold{\underline{\underline{Solution:-}}}$

$let \\ \\ a + ib = \sqrt{ - 8 - 6i} \\ squaring \: on \: both \: side \\ \\ {(a + ib)}^{2} = - 8 - 6i \\ {a}^{2} + {b}^{2} {(i)}^{2} + 2abi = - 8 - 6i \\ {a}^{2} - {b}^{2} + 2abi = - 8 - 6i \\ \\ compare \: the \: coefficients \\ we \: get \\ \\ {a}^{2} - {b}^{2} = - 8.....(2) \\ 2ab = - 6.....(1) \\ \\ take \: (1) \\ \\ 2ab = - 6 \\ ab = - 3 \\ a = \frac{ - 3}{b} ....put \: in \: (2) \\ \\ {( \frac{ - 3}{b}) }^{2} - {b}^{2} = - 8 \\ \frac{9}{ {b}^{2} } - {b}^{2} = - 8 \\ \\ take \: lcm \: \\ {9 - {b}^{4} } = - 8 {b}^{2} \\ \\ {b}^{4} - 8 {b}^{2} - 9 = 0 \\ {b}^{4} - 9 {b}^{2} + {b}^{2} - 9 = 0 \\ {b}^{2} ( {b}^{2} - 9) + 1( {b}^{2} - 9) = 0 \\ ( {b}^{2} + 1)( {b}^{2} - 9) = 0 \\ \\ from \: here \: we \: get \\ {b}^{2} = - 1(not \: possible) \\ {b}^{2} = 9 = 3 \: or \: - 3 \\ put \: value \: of \: b \: in \: eq(2) \\ {a}^{2} - 9 = - 8 \\ {a}^{2} = 1 \\ \\ so \: required \: answer \\ \sqrt{ - 8 - 6i} = 1 - 3i \\ \\ we \: take \: b = - 3 \: because \: + 3 \: \\ is \: not \: satisfying \: the \: \: condition$