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.