Question

If$\left( {x + iy} \right)\left( {2 - 3i} \right) = 4 + i$, then $\left( {x,y} \right)$ is
A) $\left( {1,\dfrac{1}{{13}}} \right)$
B) $\left( { - \dfrac{5}{{13}},\dfrac{{14}}{{13}}} \right)$
C) $\left( {\dfrac{5}{{13}},\dfrac{{14}}{{13}}} \right)$
D) $\left( { - \dfrac{5}{{13}}, - \dfrac{{14}}{{13}}} \right)$

Answer 1

Hint: The problem requires analysis and comparison after solving the Left-hand side of the equation.

Complete step by step answer:
The given expression is
$\left( {x + iy} \right)\left( {2 - 3i} \right) = 4 + i......(1)$

The expression consists of a complex number on both sides of the equal to sign and the value of x and y is to be evaluated.
First of all the LHS of the expression should be simplified
LHS
$\left( {x + iy} \right)\left( {2 - 3i} \right)$

Expanding the expression
$
  \left( {x + iy} \right)\left( {2 - 3i} \right) \\
  2x - 3xi + 2yi - 3y{i^2}......(2) \\
 $

The value of ${i^2} = 1$ , substitute it in equation (1)
$2x - 3xi + 2yi + 3y$

Now take the like terms together i.e., terms containing i and variables x and y
$(2x + 3y) + \left( { - 3x + 2y} \right)i$

Now the equation (1) becomes as
$(2x + 3y) + \left( { - 3x + 2y} \right)i = 4 + i......(3)$

On comparing the two sides of the equation (3),
$2x + 3y = 4......(i)$
$ - 3x + 2y = 1......(ii)$
In comparison, two linear equations come out in the form of x and y. Solving equation(i) and equation (ii) the values of x and y can be evaluated.

Multiply equation (i) by 3 and equation (ii) by 2,
$6x + 9y = 12......(iii)$
$ - 6x + 4y = 2......(iv)$
On adding equation (iii) and (iv) , the value of y can be calculated.
\[
  \underline
  6x + 9y = 12 \\
   - 6x + 4y = 2 \\
    \\
  0 + 13y = 14 \\
  y = \dfrac{{14}}{{13}} \\
 \]
The value of y comes out to be $y = \dfrac{{14}}{5}$ , substitute it in equation (i) to calculate the value of x.
$
  2x + 3\left( {\dfrac{{14}}{{13}}} \right) = 4 \\
  2x + \dfrac{{42}}{{13}} = 4 \\
  2x = 4 - \dfrac{{42}}{{13}} \\
  2x = \dfrac{{52 - 42}}{{13}} \\
  2x = \dfrac{{10}}{{13}} \\
  x = \dfrac{{10}}{{2 \times 13}} \\
  x = \dfrac{5}{{13}} \\
 $
The value comes out to be $x = \dfrac{5}{{13}}$

Thus, $\left( {x,y} \right) = \left( {\dfrac{5}{{13}},\dfrac{{14}}{{13}}} \right)$
Hence, option (C) is correct.

Note:
The important step in the evaluation of this problem is solving the LHS part and it’s a comparison that should be kept in mind.
 A complex number is a number that has a real and imaginary part . for instance, is a complex number in which x is the real part, and y is the imaginary part.
The value of and the value of. This should be kept in mind while evaluating the equality involving the complex numbers.