Question

Given that \[x,y\in R\], solve \[\left( x+2y \right)+i\left( 2x-3y \right)=5-4i\].

Answer 1

Hint: Given equation is of the form \[a+ib\], where a is real number and b is imaginary. Separate the real part and imaginary part from the given equation. From the 2 equations form it , solve it and get the values of x and y.

Complete step-by-step solution -

We know that a complex number is a number that can be expressed in the form \[a+ib\], where a and b are real numbers. Here ‘i’ is a solution of the equation, \[{{x}^{2}}=-1\]. No real number can satisfy the equation. So ‘i’ is called an imaginary number.
Now from the complex number \[a+ib\], a is called the real part and b is called the imaginary part. If speaking geometrically, complex numbers extend the concept of 1D number line to 2D complex plane by using the horizontal axis for the real part and the vertical axis for the imaginary part.
Now given to us, \[\left( x+2y \right)+i\left( 2x-3y \right)=5-4i\]
Now comparing this to \[a+ib\], we can separate the real part and the imaginary part.
Here, \[x+2y=5\], represents the real part and \[\left( 2x-3y \right)=-4i\] represents the imaginary port.
\[x+2y=5-(1)\]
\[i\left( 2x-3y \right)=-4i\], imaginary part cancel i on LHS and RHS.
\[\therefore 2x-3y=-4-(2)\]
Now let us solve both equation (1) and (2), multiply equation (1) with (2).
Subtract equation (1) and (2).
2x + 4y = 10
2x – 3y = -4
        7y = 14
We get, 7y = 14
\[\therefore y=\dfrac{14}{7}=2\]
Put, y = 2 in equation (1).
\[\begin{align}
  & x+\left( 2\times 2 \right)=5 \\
 & \therefore x=5-4=1 \\
\end{align}\]
Hence we got, x = 1 and y = 2.
Thus we solved the given complex equation and get value of x and y as x = 1 and y = 2.

Note: From the given expression, you should be able to classify the real and imaginary part of the complex equation, given in the form of \[a+ib.\left( 5-4i \right)\] is the conjugate of \[\left( 5+4i \right)\] i.e. it is \[\overline{z}\]. Thus, form the equations and find x, y.