Question

Given that x, y$\in$R, solve $\left( 2+3i \right){{x}^{2}}=\left( 3-2i \right)y=2x-3y+5i$

Answer 1

Hint: Given equation is of the form a + ib, where a is real part and imaginary part from the given equation. Frame 2 equation from it. Solve it and get the values of x and y.

Complete step-by-step answer:
We know that a complex number is a number that can be expressed in the from a + bi, 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 no: a + bi, 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 pest and the vertical axis for the imaginary part.
Now we have been given the complex equation,
$\left( 2+3i \right){{x}^{2}}=\left( 3-2i \right)y=2x-3y+5i$
Let us open their brackets and separate the real part and their imaginary part
$2{{x}^{2}}+3{{x}^{2}}i-3y+2yi=2x-3y+5i$
$\left( 2{{x}^{2}}-3y \right)+i\left( 3{{x}^{2}}+2y \right)=\left( 2x-3y \right)+5i$………………(i)
Thus from the above equation $2{{x}^{2}}-3y=2x-3y$ is the real part and $i\left( 3{{x}^{2}}+2y \right)=5i$ is the imaginary part.
Let us first take the real part.
$2{{x}^{2}}-3y=2x-3y$, cancel out (-3y) from both sides and simplify it
$2{{x}^{2}}=2x\Rightarrow {{x}^{2}}-x=0$ ………………… (ii)
x(x – 1) = 0, hence we get x = 0, x – 1 =0
$\therefore$ x = 0, x = 1
Now let us simplify the imaginary part
$3{{x}^{2}}+2y=5$ ……………………(iii)
Now let us put x = 0 in equation (iii)
$\begin{aligned} & 3\times 0+2y=5 \\\ & \therefore y=\dfrac{5}{2} \\\ \end{aligned}$
Thus when x = 0,$y=\dfrac{5}{2}$
Now put x = 1 in equation (iii)
$3\times {{1}^{2}}+2y=5$
2y = 5 – 3 = 2
$y=\dfrac{2}{2}=1$ $\therefore$ when x = 1, y = 1
Hence we got two solutions for the above expression, (x, y) = $\left( 0,\dfrac{5}{2} \right)$ , (1, 1)
Hence we got the required value.
Note: As there are a lot of terms, don’t get confused in between the terms of the real part and the imaginary part. We should be able to identify the real part and imaginary part of the given expression, here for ${{x}^{2}}=x$, don’t cancel it get only one case x = 1. Thus take ${{x}^{2}}-x=0$, which will give x = 0, x = 1.