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Question: If \(x,y\in R\) and \(\left( x+iy \right)\left( 3+2i \right)=1+i\) , then \(\left( x,y \right)\) is:...

If x,y∈Rx,y\in R and (x+iy)(3+2i)=1+i\left( x+iy \right)\left( 3+2i \right)=1+i , then (x,y)\left( x,y \right) is:
a). (1,15)\left( 1,\dfrac{1}{5} \right)
b). (113,113)\left( \dfrac{1}{13},\dfrac{1}{13} \right)
c). (513,113)\left( \dfrac{5}{13},\dfrac{1}{13} \right)
d). (15,15)\left( \dfrac{1}{5},\dfrac{1}{5} \right)

Explanation

Solution

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,
(x+iy)(3+2i)\left( x+iy \right)\left( 3+2i \right)
Multiplying, we get,
=(x×3)+(x×2i)+(iy×3)+(iy×2i) =3x+2xi+3yi+2yi2 \begin{aligned} & =\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{aligned}
Putting i2=1{{i}^{2}}=1 we get,
=3x+(2x+3y)i+2y(−1) =(3x−2y)+(2x+3y)i \begin{aligned} & =3x+\left( 2x+3y \right)i+2y\left( -1 \right) \\\ & =\left( 3x-2y \right)+\left( 2x+3y \right)i \\\ \end{aligned}
Now, comparing the LHS and RHS, we have,
=(3x−2y)+(2x+3y)i=1+i=\left( 3x-2y \right)+\left( 2x+3y \right)i=1+i
Thus, we have,
3x−2y=1 2x+3y=1 \begin{aligned} & 3x-2y=1 \\\ & 2x+3y=1 \\\ \end{aligned}
We should solve these two equations to get values of x and y.
Therefore,
3x−2y=1............×3 2x+3y=1............×2 \begin{aligned} & 3x-2y=1............\times 3 \\\ & 2x+3y=1............\times 2 \\\ \end{aligned}
We have,
9x−6y=3 4x+6y=2 \begin{aligned} & 9x-6y=3 \\\ & 4x+6y=2 \\\ \end{aligned}
Adding and cancelling the like terms we get,
13x=513x=5
Cross multiplication, we have.
x=513x=\dfrac{5}{13} .
Now, putting x=513x=\dfrac{5}{13} in 3x−2y=13x-2y=1we have,
3(513)−2y=13\left( \dfrac{5}{13} \right)-2y=1
Re arranging, we have,
1513−1=2y\dfrac{15}{13}-1=2y .
Taking LCM, we have,
15−1313=2y\dfrac{15-13}{13}=2y .
Thus, 2y=2132y=\dfrac{2}{13}
Therefore, y=113y=\dfrac{1}{13} .
Thus (x,y)=(513,113)\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.
(x+iy)(3+2i)=1+i\left( x+iy \right)\left( 3+2i \right)=1+i
Cross multiplying , (3+2i)\left( 3+2i \right) we have,
x+iy=1+i3+2ix+iy=\dfrac{1+i}{3+2i}
Now, multiplying and dividing by the conjugate of 3+2i3+2i , that is 3−2i3-2i , we have,
x+iy=(1+i)3+2i×3−2i3−2ix+iy=\dfrac{\left( 1+i \right)}{3+2i}\times \dfrac{3-2i}{3-2i} .
Multiplying , we have,
x+iy=(1+i)(3−2i)(3+2i)(3−2i)x+iy=\dfrac{\left( 1+i \right)\left( 3-2i \right)}{\left( 3+2i \right)\left( 3-2i \right)}
Which is,
x+iy=(1×3)+(1×−2i)+(i×3)+(i×−2i)(3×3)+(3×−2i)+(2i×3)+(2i×−2i) x+iy=3−2i+3i−2i29−6i+6i−4i2 \begin{aligned} & 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{aligned}
Substituting i2=−1{{i}^{2}}=-1 , we get,
x+iy=3+i−2(−1)9−4(−1) x+iy=3+i+29+4 x+iy=5+i13. \begin{aligned} & 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{aligned}
Splitting the terms,
x+iy=513+113ix+iy=\dfrac{5}{13}+\dfrac{1}{13}i
Thus, (x,y)=(513,113)\left( x,y \right)=\left( \dfrac{5}{13},\dfrac{1}{13} \right) .