Question

If the conjugate of $\left( {x + iy} \right)\left( {1 - 2i} \right)$ is $\left( {1 + i} \right)$, then;
$\left( A \right)x - iy = \dfrac{{1 - i}}{{1 - 2i}}$
$\left( B \right)x + iy = \dfrac{{1 - i}}{{1 - 2i}}$
$\left( C \right)x - iy = \dfrac{{1 - i}}{{1 + 2i}}$
$\left( D \right)x - iy = \dfrac{{1 - i}}{{1 + i}}$

Answer 1

To solve this question, we should be familiar with the concept of the conjugate of a complex number. If $a + ib$ is the complex number where $a$ represents the real part and $b$ is the imaginary part , then the complex conjugate of $a + ib = a - ib$. Each of the two i.e. the conjugate and the complex number have equal and same real parts but their imaginary parts differ in sign.

Complete step by step answer:
The first step is to find out the complex conjugate of the given complex number. The given complex number is $\left( {x + iy} \right)\left( {1 - 2i} \right)$,
A complex number is generally denoted by z;
$\Rightarrow z = \left( {x + iy} \right)\left( {1 - 2i} \right)$
Let us simplify this first,
$\Rightarrow z = x - 2xi + iy - 2\left( {{i^2}} \right)y$
We know that ${i^2} = - 1$ , the squaring of an imaginary number gives us a negative value.
$\Rightarrow z = x - 2xi + iy + 2y$
Rearranging the real parts and the imaginary parts;
$\Rightarrow z = \left( {x + 2y} \right) + i\left( {y - 2x} \right)$
According to the definition discussed above, the complex conjugate of $z$ will be $\bar z$which equals to;
$\Rightarrow \bar z = \left( {x + 2y} \right) - i\left( {y - 2x} \right)$
$\Rightarrow \bar z = \left( {x + 2y} \right) + i\left( {2x - y} \right)$ $......\left( 1 \right)$
As stated in the given question;
$\Rightarrow \bar z = 1 + i$ $......\left( 2 \right)$
Comparing equation $\left( 1 \right)$ and $\left( 2 \right)$, we get;
$\Rightarrow \bar z = \left( {x + 2y} \right) + i\left( {2x - y} \right) = 1 + i$
Comparing the real part and imaginary part of the above equation we get;
$\Rightarrow x + 2y = 1$ $......\left( 3 \right)$
$\Rightarrow 2x - y = 1$ $......\left( 4 \right)$
Multiplying equation $\left( 4 \right)$ by $2$, we get;
$\Rightarrow 4x - 2y = 2$ $......\left( 5 \right)$
Now adding equation $\left( 3 \right)$ and equation $\left( 5 \right)$ ;
$\Rightarrow x + 2y + 4x - 2y = 1 + 2$
$\Rightarrow 5x = 3$
$\Rightarrow x = \dfrac{3}{5}$
Now substitute the value of $x$ in equation $\left( 3 \right)$, to get the value of $y$;
$\Rightarrow \dfrac{3}{5} + 2y = 1$
$\Rightarrow 2y = 1 - \dfrac{3}{5}$
$\Rightarrow 2y = \dfrac{2}{5}$
$\therefore y = \dfrac{1}{5}$
Since we have found the values of $x$ and $y$, we can write the complex number as;
$\Rightarrow x + iy = \dfrac{3}{5} + \dfrac{i}{5}$
We can also represent it as;
$\Rightarrow x + iy = \dfrac{1}{5}\left( {3 + i} \right)$
According to the given question, we have to check which option satisfies our condition; but we do not have to check every option. Since the complex number has a positive imaginary part , we will only check option $\left( 2 \right)$. In option $\left( 2 \right)$ $x + iy = \dfrac{{1 - i}}{{1 - 2i}}$ , Now we have to simplify it to check if it satisfies our solution or not;
To simplify it we have to rationalize the denominator by taking its conjugate. The denominator is $1 - 2i$,
so its conjugate will be $\left( {1 + 2i} \right)$.
$\Rightarrow x + iy = \dfrac{{1 - i}}{{1 - 2i}} \times \dfrac{{1 + 2i}}{{1 + 2i}}$
$\Rightarrow \dfrac{{\left( {1 - i} \right)\left( {1 + 2i} \right)}}{{{1^2} - {{\left( {2i} \right)}^2}}} = \dfrac{{1 + 2i - i - 2{i^2}}}{{1 - \left( { - 4} \right)}}$
$\Rightarrow x + iy = \dfrac{{3 + i}}{5}$
Or $x + iy = \dfrac{1}{5}\left( {3 + i} \right)$
Therefore option $\left( 2 \right)$, satisfies the solution.
The correct answer for this question is option $\left( B \right)$ i.e. $x + iy = \dfrac{{1 - i}}{{1 + 2i}}$.

Note:
To solve this question we need two concepts. The first one is to find the conjugate of a complex number. The second one is rationalizing the denominator with its conjugate. It is difficult to deal with imaginary numbers as compared to real numbers while doing calculations. Finding the conjugate helps to rationalize the complex numbers and also eases the calculation.