Question

Find the square root of 8 – 6i. ( ${i^2} = - 1$ )

Answer 1

Hint: Assign a complex number x + iy to the square root of 8 – 6i and then square both sides and simplify to get two equations in two unknowns. Solve the equations to get the square root of 8 – 6i.

Complete step-by-step answer:
Let us assign a complex number x + iy to the square root of 8 – 6i such that both x and y are real numbers. Then, we have:
$x + iy = \sqrt {8 - 6i}$
Squaring both sides, we have:
${\left( {x + iy} \right)^2} = 8 - 6i$
Simplifying the left-hand side of the expression, we get:
${x^2} + 2ixy - {y^2} = 8 - 6i$
Grouping the terms and comparing on both the sides of the equation, we get:
$({x^2} - {y^2}) + 2ixy = 8 - 6i$
${x^2} - {y^2} = 8.............(1)$
$2xy = - 6.............(2)$
Hence, we have two equations (1) and (2) in two unknowns x and y.
From the second equation, we have:
$xy = - 3............(3)$
Squaring this equation, we have:
${x^2}{y^2} = 9............(4)$
From equation (1), we have:
${x^2} = {y^2} + 8.............(5)$
Substituting equation (5) in equation (4), we have:
$({y^2} + 8){y^2} = 9$
Let us take ${y^2} = \omega$, then we have:
$(\omega + 8)\omega = 9$
Simplifying, we have:
${\omega ^2} + 8\omega - 9 = 0$
We can write $8\omega$ as the sum of $9\omega$ and $- \omega$, then, we have:
${\omega ^2} + 9\omega - \omega - 9 = 0$
Simplifying, we have:
$\omega (\omega + 9) - 1(\omega + 9) = 0$
$(\omega - 1)(\omega + 9) = 0$
$\omega = 1;\omega = - 9$
y is a real number, hence, its square can’t be negative, hence, we have:
$\omega = 1$
${y^2} = 1$
$y = \pm 1...........(6)$
For y = 1, from equation (3), we have:
$x = - 3$
Hence, one root is $- 3 + i$.
For y = – 1, from equation (3), we have:
$x( - 1) = - 3$
$x = 3$
Hence, another root is $3 - i$.
Hence, the square roots of 8 – 6i are – 3 + i and 3 – i.

Note: You can also write 8 – 6i as (9 – 2.3i – 1) and use the formula ${(a - b)^2} = {a^2} - 2ab + {b^2}$ to simplify the expression and take the square root to find the roots. Here we have rejected y as x and y should be real numbers, but here y is in the form of i.