Question

If the conjugate of $\left( {x + iy} \right)\left( {1 - 2i} \right)$ be $1 + i$, then:
(A) $x = \dfrac{3}{5}$
(B) $y = \dfrac{1}{5}$
(C) $x + iy = \dfrac{{1 - i}}{{1 - 2i}}$
(D) $x - iy = \dfrac{{1 - i}}{{1 + 2i}}$

Answer 1

Hint: First, we have to find the product of the given terms. Then find the conjugate of the complex number obtained. Then equate its real and imaginary parts to the given value. Finally we have to find the value of $x$ and $y$. Also, conclude the correct answer.

Formula used:
Let $z = a + ib$ be a complex number. Then the conjugate of $z$ is denoted by $\overline z $ and is equal to $a - ib$.
Thus, $z = a + ib \Rightarrow \overline z = a - ib$.

Complete step-by-step solution:
Let us consider the given as $z = \left( {x + iy} \right)\left( {1 - 2i} \right)$.
Simplify $z$ by finding the product of $\left( {x + iy} \right)$ and $\left( {1 - 2i} \right)$.
Multiply $x$ with $\left( {1 - 2i} \right)$ and $iy$ with $\left( {1 - 2i} \right)$ and add them.
$z = x\left( {1 - 2i} \right) + iy\left( {1 - 2i} \right)$
Multiply $x$ and $iy$ with each term in bracket
$z = x\left( 1 \right) + x\left( { - 2i} \right) + iy\left( 1 \right) + iy\left( { - 2i} \right)$
On multiply the bracket term and we get
$z = x + - 2ix + iy - 2{i^2}y$
Putting the value of ${i^2}$ in the above equation
As,$i = \sqrt { - 1} $
On squaring both sides we get,
$ \Rightarrow {i^2} = - 1$
So, after putting${i^2} = - 1$, $z$ becomes
$z = x + - 2ix + iy + 2y$
It can be rewritten as $z = x + 2y + i\left( {y - 2x} \right)$.
Now we have to calculate the conjugate of $z$ using the formula for conjugate of complex number.
As $z = x + 2y + i\left( {y - 2x} \right)$.
Replace $i$ with $ - i$, we get
$\overline z = x + 2y - i\left( {y - 2x} \right)$
It can be rewritten as$\overline z = x + 2y + i\left( {2x - y} \right)$.
Compare $\overline z $ with $1 + i$ by equating real and imaginary parts of both equations.
$\overline z = 1 + i$
⇒ $x + 2y + i\left( {2x - y} \right) = 1 + i$
Here we have to equating the real and the imaginary parts from both sides, we get
$x + 2y = 1....\left( 1 \right)$
$2x - y = 1....\left( 2 \right)$
Solving equations (1) and (2) to get the value of $x$ and $y$.
Multiply equation (2) by 2, we get
$ \Rightarrow 4x - 2y = 2...\left( 3 \right)$
Adding equations (1) and (3), we get
$ \Rightarrow x + 2y + 4x - 2y = 1 + 2$
On adding and subtract the same variables, we get
$ \Rightarrow 5x = 3$
On dividing \[5\] on both sides we get,
$ \Rightarrow x = \dfrac{3}{5}$
Put the value of $x$ in equation (2).
$ \Rightarrow 2 \times \dfrac{3}{5} - y = 1$
On multiply the term, we get
$ \Rightarrow \dfrac{6}{5} - y = 1$
Taking y as RHS and remaining as LHS we get
$ \Rightarrow y = \dfrac{6}{5} - 1$
Taking LCM,
$ \Rightarrow y = \dfrac{1}{5}$

Thus, (A) $x = \dfrac{3}{5}$ and (B) $y = \dfrac{1}{5}$ are correct options.

Note: If $a$, $b$ are two real numbers, then a number of the form $a + ib$ is called a complex number.
If $z = a + ib$ is a complex number, then ‘$a$’ is called the real part of $z$ and ‘$b$’ is known as the imaginary part of$z$.
Now, we take any positive real number $a$, we have$\sqrt { - a} = \sqrt { - 1 \times a} = \sqrt { - 1} \times \sqrt a = i\sqrt a $.
If ${z_1} = a + bi$ and ${z_2} = c + di$ be two complex numbers. Then the multiplication of ${z_1}$ with ${z_2}$ is denoted by ${z_1} \cdot {z_2}$ and is defined as the complex number $\left( {ac - bd} \right) + \left( {ad + bc} \right)i$.
Thus, ${z_1} \times {z_2} = \left( {a + bi} \right) \times \left( {c + di} \right)$
$ = a\left( {c + di} \right) + bi\left( {c + di} \right)$
$ = ac + adi + bci + bd{i^2}$ $\left[ {\because {i^2} = - 1} \right]$
$ = \left( {ac - bd} \right) + \left( {ad + bc} \right)i$