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Question: Given that x, y\(\in \)R, solve \(4{{x}^{2}}+3xy+\left( 2xy-3{{x}^{2}} \right)i=4{{y}^{2}}-\dfrac{{{...

Given that x, y∈\in R, solve 4x2+3xy+(2xy−3x2)i=4y2−x22+(3xy−2y2)i4{{x}^{2}}+3xy+\left( 2xy-3{{x}^{2}} \right)i=4{{y}^{2}}-\dfrac{{{x}^{2}}}{2}+\left( 3xy-2{{y}^{2}} \right)i.

Explanation

Solution

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 no. is a no. that can be expressed in the from a + bi, where a and b are real numbers. Here ‘i’ is a solution of the equation x2=−1{{x}^{2}}=-1 no real no: can satisfy the equation. So ‘i’ is called an imaginary number.
Now from the complex no: a + bi, a is called real part and b is called the imaginary pest. If speaking geometrically, complex no: extends 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 given to us the expression,
(4x2+3xy)+(2xy−3x2)i=(4y2−x22)+(3xy−2y2)i\left( 4{{x}^{2}}+3xy \right)+\left( 2xy-3{{x}^{2}} \right)i=\left( 4{{y}^{2}}-\dfrac{{{x}^{2}}}{2} \right)+\left( 3xy-2{{y}^{2}} \right)i
Now let us separate the real part and their imaginary part of the above given expression.
Real part ⇒(4x2+3xy)=4y2−x22\Rightarrow \left( 4{{x}^{2}}+3xy \right)=4{{y}^{2}}-\dfrac{{{x}^{2}}}{2}
Imaginary part ⇒2xy−3x2=3xy−2y2\Rightarrow 2xy-3{{x}^{2}}=3xy-2{{y}^{2}}
Now let us first take the real part,
4x2+3xy=4y2−x22 4x2−4y2+3xy+x22=0 x2(4+12)−4y2+3xy=0 x2(8+12)−4y2+3xy=0 \begin{aligned} & 4{{x}^{2}}+3xy=4{{y}^{2}}-\dfrac{{{x}^{2}}}{2} \\\ & 4{{x}^{2}}-4{{y}^{2}}+3xy+\dfrac{{{x}^{2}}}{2}=0 \\\ & {{x}^{2}}\left( 4+\dfrac{1}{2} \right)-4{{y}^{2}}+3xy=0 \\\ & {{x}^{2}}\left( \dfrac{8+1}{2} \right)-4{{y}^{2}}+3xy=0 \\\ \end{aligned}
9x22−4y2+3xy=0\dfrac{9{{x}^{2}}}{2}-4{{y}^{2}}+3xy=0…………………..(i)
From the imaginary part, let's simplify it.
2xy−3x2=3xy−2y2 3x2−2y2=−3xy=+2xy 3x2−2y2+xy=0 \begin{aligned} & 2xy-3{{x}^{2}}=3xy-2{{y}^{2}} \\\ & 3{{x}^{2}}-2{{y}^{2}}=-3xy=+2xy \\\ & 3{{x}^{2}}-2{{y}^{2}}+xy=0 \\\ \end{aligned}
i.e., xy=2y2−3x2xy=2{{y}^{2}}-3{{x}^{2}}…………….(ii)
Now put the volume of xy in equation (i)

&xy;=2{{y}^{2}}-3{{x}^{2}} \\\ & \dfrac{9{{x}^{2}}}{2}-4{{y}^{2}}-3xy=0 \\\ & \dfrac{9{{x}^{2}}}{2}-4{{y}^{2}}-3\left( 2{{y}^{2}}-3{{x}^{2}} \right)=0 \\\ \end{aligned}$$ Let us simplify it $\begin{aligned} & \dfrac{9{{x}^{2}}}{2}-4{{y}^{2}}+6{{y}^{2}}-9{{x}^{2}}=0 \\\ & \dfrac{9{{x}^{2}}}{2}-9{{x}^{2}}+2{{y}^{2}}=0 \\\ & 9{{x}^{2}}\left[ \dfrac{1}{2}-1 \right]+2{{y}^{2}}=0 \\\ & 9{{x}^{2}}\left[ \dfrac{1-2}{2} \right]+2{{y}^{2}}=0 \\\ & \dfrac{-9{{x}^{2}}}{2}=-2{{y}^{2}}\Rightarrow 9{{x}^{2}}=4{{y}^{2}} \\\ \end{aligned}$ $\dfrac{{{x}^{2}}}{{{y}^{2}}}=\dfrac{9}{4}$, taking square root on both sides, we get $\dfrac{x}{y}=\sqrt{\dfrac{9}{4}}=\dfrac{3}{2},\dfrac{x}{y}=\dfrac{3}{2}$ Now multiply the numerator and denominator by k, where k is the proportionality constant i.e., $\dfrac{x}{y}=\dfrac{3k}{2k}$ Where x = k, $y=\dfrac{3}{2}k$ hence, we got (x, y) $\to \left( k,\dfrac{3}{2}k \right)$ Thus, we got the required values. Note: As we got the answer in the form of a radio, use a proportionality constant l. don’t leave the answers such and write (x, y)$\to $ (3, 2), which is wrong.