Question

If $x,y\in R$ and $\left( x+iy \right)\left( 3+2i \right)=1+i$ , then $\left( x,y \right)$ is:
a). $\left( 1,\dfrac{1}{5} \right)$
b). $\left( \dfrac{1}{13},\dfrac{1}{13} \right)$
c). $\left( \dfrac{5}{13},\dfrac{1}{13} \right)$
d). $\left( \dfrac{1}{5},\dfrac{1}{5} \right)$

Answer 1

Hint: considering the LHS part, first we will multiply both the terms and convert it into a linear a+bi term. Then we will compare the LHS and RHS. Doing so, we will get two conditions in x and y. Solving the two simultaneous linear equations we get the value of x and y.

Complete step-by-step solution -
Considering LHS, we have,
$\left( x+iy \right)\left( 3+2i \right)$
Multiplying, we get,
$\begin{align}
  & =\left( x\times 3 \right)+\left( x\times 2i \right)+\left( iy\times 3 \right)+\left( iy\times 2i \right) \\
 & =3x+2xi+3yi+2y{{i}^{2}} \\
\end{align}$
Putting ${{i}^{2}}=1$ we get,
$\begin{align}
  & =3x+\left( 2x+3y \right)i+2y\left( -1 \right) \\
 & =\left( 3x-2y \right)+\left( 2x+3y \right)i \\
\end{align}$
Now, comparing the LHS and RHS, we have,
$=\left( 3x-2y \right)+\left( 2x+3y \right)i=1+i$
Thus, we have,
$\begin{align}
  & 3x-2y=1 \\
 & 2x+3y=1 \\
\end{align}$
We should solve these two equations to get values of x and y.
Therefore,
$\begin{align}
  & 3x-2y=1............\times 3 \\
 & 2x+3y=1............\times 2 \\
\end{align}$
We have,
$\begin{align}
  & 9x-6y=3 \\
 & 4x+6y=2 \\
\end{align}$
Adding and cancelling the like terms we get,
$13x=5$
Cross multiplication, we have.
$x=\dfrac{5}{13}$ .
Now, putting $x=\dfrac{5}{13}$ in $3x-2y=1$we have,
\[3\left( \dfrac{5}{13} \right)-2y=1\]
Re arranging, we have,
$\dfrac{15}{13}-1=2y$ .
Taking LCM, we have,
$\dfrac{15-13}{13}=2y$ .
Thus, $2y=\dfrac{2}{13}$
Therefore, $y=\dfrac{1}{13}$ .
Thus \[\left( x,y \right)=\left( \dfrac{5}{13},\dfrac{1}{13} \right)\]
Therefore, option (c ) is correct.

Note: There is an alternate method to solve this. Consider the given equation.
$\left( x+iy \right)\left( 3+2i \right)=1+i$
Cross multiplying , $\left( 3+2i \right)$ we have,
\[x+iy=\dfrac{1+i}{3+2i}\]
Now, multiplying and dividing by the conjugate of $3+2i$ , that is $3-2i$ , we have,
$x+iy=\dfrac{\left( 1+i \right)}{3+2i}\times \dfrac{3-2i}{3-2i}$ .
Multiplying , we have,
$x+iy=\dfrac{\left( 1+i \right)\left( 3-2i \right)}{\left( 3+2i \right)\left( 3-2i \right)}$
Which is,
$\begin{align}
  & x+iy=\dfrac{\left( 1\times 3 \right)+\left( 1\times -2i \right)+\left( i\times 3 \right)+\left( i\times -2i \right)}{\left( 3\times 3 \right)+\left( 3\times -2i \right)+\left( 2i\times 3 \right)+\left( 2i\times -2i \right)} \\
 & x+iy=\dfrac{3-2i+3i-2{{i}^{2}}}{9-6i+6i-4{{i}^{2}}} \\
\end{align}$
Substituting ${{i}^{2}}=-1$ , we get,
$\begin{align}
  & x+iy=\dfrac{3+i-2\left( -1 \right)}{9-4\left( -1 \right)} \\
 & x+iy=\dfrac{3+i+2}{9+4} \\
 & x+iy=\dfrac{5+i}{13}. \\
\end{align}$
Splitting the terms,
$x+iy=\dfrac{5}{13}+\dfrac{1}{13}i$
Thus, $\left( x,y \right)=\left( \dfrac{5}{13},\dfrac{1}{13} \right)$ .