Question

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

Answer 1

Hint: Given equation is of the form a + ib, where a is real pest 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 pest.
Now we have been given the complex equation,
$\begin{aligned} & {{x}^{2}}-{{y}^{2}}-i\left( 2x+y \right)=2i \\\ & \left( {{x}^{2}}-{{y}^{2}} \right)-i\left( 2x+y \right)=0+2i \\\ \end{aligned}$
From this we can say that ${{x}^{2}}-{{y}^{2}}=0$is the real part and -i(2x + y) = 2i is the imaginary part.
We know that ${{x}^{2}}-{{y}^{2}}=\left( x-y \right)\left( x+y \right)$
${{x}^{2}}-{{y}^{2}}=0$ …………………… (i)
(x – y) (x + y) = 0
Thus x – y = 0, we get x = y
Similarly, from x + y = 0, we get x = - y
Hence there are 2 cases.
Now i(2x + y) = 2i is the imaginary part. Thus, we get the 2nd equation as
2x + y = +2……………………… (ii)
Now put x = y in equation (ii)
2x + y = +2
2y + y = +2$\Rightarrow$ 3y = +2,$y=\dfrac{+2}{3}$
Now put x = -y
2(-y) + y = +2
-2y + y = +2$\Rightarrow$ y = -2
Hence, when x = y, $y=\dfrac{+2}{3}$
When x = -y, y = -2
Thus (x, y) = $\left( \dfrac{2}{3},\dfrac{2}{3} \right)$, (2, -2)
Hence, we got 2 values of (x, y) $\to$ $\left( \dfrac{2}{3},\dfrac{2}{3} \right)$ , (2, -2)

Note: From the real part ${{x}^{2}}-{{y}^{2}}=0$, we might ${{x}^{2}}={{y}^{2}}$and take their square root to get x = y, which is only one case. Thus expand it with formula to get two case x = y, x = -y