Question

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

Answer 1

Hint: We first simplify the equation in the complex expression. We take the multiplication of the denominators. We also use identity value of ${{i}^{2}}=-1,{{i}^{3}}=-i,{{i}^{4}}=1$ for simplification. We equate the coefficients to find the value of $x$ and $y$.

Complete step by step answer:
We need to simplify the equation of $\dfrac{\left( 1+i \right)x-2i}{3+i}+\dfrac{\left( 2-3i \right)y+i}{3-i}$.
We take the LCM of the denominators as the multiplication.
So, $\dfrac{\left( 1+i \right)x-2i}{3+i}+\dfrac{\left( 2-3i \right)y+i}{3-i}=\dfrac{\left( 3-i \right)\left[ \left( 1+i \right)x-2i \right]+\left( 3+i \right)\left[ \left( 2-3i \right)y+i \right]}{\left( 3+i \right)\left( 3-i \right)}$.
We know the identity theorem $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$.
Here we have a complex number and we denote that as \[\sqrt{-1}=i\]. We have the relations for imaginary $i$ where ${{i}^{2}}=-1,{{i}^{3}}=-i,{{i}^{4}}=1$.
We get $\left( 3+i \right)\left( 3-i \right)={{3}^{2}}-{{i}^{2}}=9+1=10$.
Now we complete the numerator. The first one gives
$\begin{align}
  & \left( 3-i \right)\left[ \left( 1+i \right)x-2i \right] \\
 & =3x\left( 1+i \right)+2{{i}^{2}}-6i-ix\left( 1+i \right) \\
 & =4x-2+i\left( 2x-6 \right) \\
\end{align}$
The second one gives
$\begin{align}
  & \left( 3+i \right)\left[ \left( 2-3i \right)y+i \right] \\
 & =3y\left( 2-3i \right)+{{i}^{2}}+3i+iy\left( 2-3i \right) \\
 & =9y-1+i\left( 3-7y \right) \\
\end{align}$
The final simplification is
\[\dfrac{\left( 1+i \right)x-2i}{3+i}+\dfrac{\left( 2-3i \right)y+i}{3-i}=\dfrac{4x-2+i\left( 2x-6 \right)+9y-1+i\left( 3-7y \right)}{10}\]
On equating with $i$ we get
\[\begin{align}
  & \dfrac{4x-2+i\left( 2x-6 \right)+9y-1+i\left( 3-7y \right)}{10}=i \\
 & \Rightarrow \left( 4x+9y-3 \right)+i\left( 2x-7y-3 \right)=10i \\
\end{align}\]
Equating coefficients, we get \[\left( 4x+9y-3 \right)=0,\left( 2x-7y-3 \right)=10\] which gives two equations
\[\begin{align}
  & 4x+9y=3.......(i) \\
 & 2x-7y=13..........(ii) \\
\end{align}\]
We multiply 2 to the both sides of the second equation and get
$\begin{align}
  & 2\times \left( 2x-7y \right)=13\times 2 \\
 & \Rightarrow 4x-14y=26 \\
\end{align}$
We take the equation as $4x-14y=26.....(iii)$.
Now we subtract the equation (iii) from equation (i) and get
$\begin{align}
  & \left( 4x+9y \right)-\left( 4x-14y \right)=3-26 \\
 & \Rightarrow 23y=-23 \\
 & \therefore y=-1 \\
\end{align}$.
Now putting the value in the equation \[4x+9y=3.......(i)\], we get
$\begin{align}
  & 4x+9y=3 \\
 & \Rightarrow x=\dfrac{3-9\times \left( -1 \right)}{4}=3 \\
\end{align}$.
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.