Solveeit Logo
Login

Question

Question: The value of \(x\) and \(y\) satisfying the equation \(\dfrac{\left( 1+i \right)x-2i}{3+i}+\dfrac{\l...

The value of xx and yy satisfying the equation (1+i)x−2i3+i+(2−3i)y+i3−i=i\dfrac{\left( 1+i \right)x-2i}{3+i}+\dfrac{\left( 2-3i \right)y+i}{3-i}=i are
A. x=−1,y=3x=-1,y=3
B. x=3,y=−1x=3,y=-1
C. x=0,y=1x=0,y=1
D. x=1,y=0x=1,y=0

Explanation

Solution

We first simplify the equation in the complex expression. We take the multiplication of the denominators. We also use identity value of i2=−1,i3=−i,i4=1{{i}^{2}}=-1,{{i}^{3}}=-i,{{i}^{4}}=1 for simplification. We equate the coefficients to find the value of xx and yy.

Complete step by step answer:
We need to simplify the equation of (1+i)x−2i3+i+(2−3i)y+i3−i\dfrac{\left( 1+i \right)x-2i}{3+i}+\dfrac{\left( 2-3i \right)y+i}{3-i}.
We take the LCM of the denominators as the multiplication.
So, (1+i)x−2i3+i+(2−3i)y+i3−i=(3−i)[(1+i)x−2i]+(3+i)[(2−3i)y+i](3+i)(3−i)\dfrac{\left( 1+i \right)x-2i}{3+i}+\dfrac{\left( 2-3i \right)y+i}{3-i}=\dfrac{\left( 3-i \right)\left[ \left( 1+i \right)x-2i \right]+\left( 3+i \right)\left[ \left( 2-3i \right)y+i \right]}{\left( 3+i \right)\left( 3-i \right)}.
We know the identity theorem (a+b)(a−b)=a2−b2\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}.
Here we have a complex number and we denote that as −1=i\sqrt{-1}=i. We have the relations for imaginary ii where i2=−1,i3=−i,i4=1{{i}^{2}}=-1,{{i}^{3}}=-i,{{i}^{4}}=1.
We get (3+i)(3−i)=32−i2=9+1=10\left( 3+i \right)\left( 3-i \right)={{3}^{2}}-{{i}^{2}}=9+1=10.
Now we complete the numerator. The first one gives
(3−i)[(1+i)x−2i] =3x(1+i)+2i2−6i−ix(1+i) =4x−2+i(2x−6) \begin{aligned} & \left( 3-i \right)\left[ \left( 1+i \right)x-2i \right] \\\ & =3x\left( 1+i \right)+2{{i}^{2}}-6i-ix\left( 1+i \right) \\\ & =4x-2+i\left( 2x-6 \right) \\\ \end{aligned}
The second one gives
(3+i)[(2−3i)y+i] =3y(2−3i)+i2+3i+iy(2−3i) =9y−1+i(3−7y) \begin{aligned} & \left( 3+i \right)\left[ \left( 2-3i \right)y+i \right] \\\ & =3y\left( 2-3i \right)+{{i}^{2}}+3i+iy\left( 2-3i \right) \\\ & =9y-1+i\left( 3-7y \right) \\\ \end{aligned}
The final simplification is
(1+i)x−2i3+i+(2−3i)y+i3−i=4x−2+i(2x−6)+9y−1+i(3−7y)10\dfrac{\left( 1+i \right)x-2i}{3+i}+\dfrac{\left( 2-3i \right)y+i}{3-i}=\dfrac{4x-2+i\left( 2x-6 \right)+9y-1+i\left( 3-7y \right)}{10}
On equating with ii we get

& \dfrac{4x-2+i\left( 2x-6 \right)+9y-1+i\left( 3-7y \right)}{10}=i \\\ & \Rightarrow \left( 4x+9y-3 \right)+i\left( 2x-7y-3 \right)=10i \\\ \end{aligned}$$ Equating coefficients, we get $$\left( 4x+9y-3 \right)=0,\left( 2x-7y-3 \right)=10$$ which gives two equations $$\begin{aligned} & 4x+9y=3.......(i) \\\ & 2x-7y=13..........(ii) \\\ \end{aligned}$$ We multiply 2 to the both sides of the second equation and get $\begin{aligned} & 2\times \left( 2x-7y \right)=13\times 2 \\\ & \Rightarrow 4x-14y=26 \\\ \end{aligned}$ We take the equation as $4x-14y=26.....(iii)$. Now we subtract the equation (iii) from equation (i) and get $\begin{aligned} & \left( 4x+9y \right)-\left( 4x-14y \right)=3-26 \\\ & \Rightarrow 23y=-23 \\\ & \therefore y=-1 \\\ \end{aligned}$. Now putting the value in the equation $$4x+9y=3.......(i)$$, we get $\begin{aligned} & 4x+9y=3 \\\ & \Rightarrow x=\dfrac{3-9\times \left( -1 \right)}{4}=3 \\\ \end{aligned}$. Therefore, the values are $x=3,y=-1$. **So, the correct answer is “Option B”.** **Note:** The integer power value of every eve number on $i$ will give the real number. Odd numbers of power value give back the imaginary part. The relation ${{i}^{2}}=-1,{{i}^{3}}=-i,{{i}^{4}}=1$ can also be represented as the unit circle on the complex plane.