Question

Find real numbers x and y if $\left( {x - iy} \right)\left( {3 + 5i} \right)$ is a conjugate of $- 6 - 24i.1 + i.1 - i$

Answer 1

First Find the product of the terms given and then its conjugate. Then equating the real part and complex part coefficient and then solving them will give us the required solution.

Complete step by step answer:

Given that,
$- 6 - 24i.1 + i.1 - i$

$$\Rightarrow \left( { - 6 - 24i} \right)\left( {1 - {i^2}} \right) \\\ \Rightarrow \left( { - 6 - 24i} \right)\left( {1 - \left( { - 1} \right)} \right) \\\ \Rightarrow \left( { - 6 - 24i} \right) \times 2 \\\ \Rightarrow - 12 + 48i \\\$$

Conjugate of $- 12 - 48i$ is $- 12 + 48i$.
Now,
$\left( {x - iy} \right)\left( {3 + 5i} \right)$

$$\Rightarrow 3x + 5xi - 3yi + 5y \\\ \Rightarrow \left( {3x + 5y} \right) + i\left( {5x - 3y} \right) \\\$$

We have both in $a + bi$ form.
So we get,
$3x + 5y = - 12$.......equation1
$5x - 3y = 48$......equation2
Solving these equations we will get values of x and y.
Multiply equation1 by 3 and equation2 by 5
$\Rightarrow (3x + 5y = - 12) \times 3$
$\Rightarrow 9x + 15y = - 36$….equa1.1
Now for equation2
$\Rightarrow (5x - 3y = 48) \times 5$
$\Rightarrow 25x - 15y = 240$…..equa2.1
Adding equa1.1 and equa2.1

$$\Rightarrow 34x = 204 \\\ \Rightarrow x = \dfrac{{204}}{{34}} \\\ \Rightarrow x = 6 \\\$$

Hence to find value of y put this value of x in any on eof the equations above
Putting it in equa2.1

$$\Rightarrow 25 \times 6 - 15y = 240 \\\ \Rightarrow 150 - 15y = 240 \\\ \Rightarrow 15y = 150 - 240 \\\ \Rightarrow y = \dfrac{{ - 90}}{{15}} \\\ \Rightarrow y = - 6 \\\$$

Thus real numbers x=6 and y=-6.

Note: Complex conjugate of a+bi is a-bi. Don’t forget to take the complex conjugate of that number. Value of ${i^2}$=-1.
Additional information: Complex numbers are of the form a+bi, where a and b are real numbers and i is the imaginary unit.
A real number if to be treated as an imaginary number it is written as a+0i.
If a number is to be treated as purely imaginary then it is written as 0+bi.