Solveeit Logo
Login

Question

Question: The real values of x and y, if \[\dfrac{\left( 1+i \right)x-2i}{3+i}+\dfrac{\left( 2-3i \right)y+i}{...

The real values of x and y, if (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 λ\lambda and μ\mu respectively, then λ−μ\lambda -\mu is

Explanation

Solution

In this type of question we have to use the concept of complex numbers. We know that in general a complex number is expressed as z=x+iyz=x+iy where xx is the real part and yy is the imaginary part of the complex number zz. Also we know about the value of ii that is i=−1i=\sqrt{-1} and hence, i2=−1{{i}^{2}}=-1. Here, first we separate out the real and imaginary parts of the both numerator and then by performing cross multiplication we can simplify it further to obtain the required result.

Complete step-by-step solution:
Now we have to find λ−μ\lambda -\mu if (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 and the real values of xx and yyare represented by λ\lambda and μ\mu respectively.
Let us consider,
⇒(1+i)x−2i3+i+(2−3i)y+i3−i=i\Rightarrow \dfrac{\left( 1+i \right)x-2i}{3+i}+\dfrac{\left( 2-3i \right)y+i}{3-i}=i
By simplifying the numerator we can write
⇒x+ix−2i(3+i)+2y−3iy+i(3−i)=i\Rightarrow \dfrac{x+ix-2i}{\left( 3+i \right)}+\dfrac{2y-3iy+i}{\left( 3-i \right)}=i
Now, we will separate real and imaginary part
⇒x+i(x−2)(3+i)+2y−i(3y−1)(3−i)=i\Rightarrow \dfrac{x+i\left( x-2 \right)}{\left( 3+i \right)}+\dfrac{2y-i\left( 3y-1 \right)}{\left( 3-i \right)}=i
By performing cross multiplication we get,
⇒(3−i)(x+i(x−2))+(3+i)(2y−i(3y−1))(3+i)(3−i)=i\Rightarrow \dfrac{\left( 3-i \right)\left( x+i\left( x-2 \right) \right)+\left( 3+i \right)\left( 2y-i\left( 3y-1 \right) \right)}{\left( 3+i \right)\left( 3-i \right)}=i
We know that, (a+b)(a−b)=(a2−b2)\left( a+b \right)\left( a-b \right)=\left( {{a}^{2}}-{{b}^{2}} \right)
⇒3x+3i(x−2)−ix−i2(x−2)+6y−3i(3y−1)+2iy−i2(3y−1)(9−i2)=i\Rightarrow \dfrac{3x+3i\left( x-2 \right)-ix-{{i}^{2}}\left( x-2 \right)+6y-3i\left( 3y-1 \right)+2iy-{{i}^{2}}\left( 3y-1 \right)}{\left( 9-{{i}^{2}} \right)}=i
Now by substituting i2=−1{{i}^{2}}=-1 and simplifying we get,

& \Rightarrow \dfrac{3x+3i\left( x-2 \right)-ix-\left( -1 \right)\left( x-2 \right)+6y-3i\left( 3y-1 \right)+2iy-\left( -1 \right)\left( 3y-1 \right)}{\left( 9-\left( -1 \right) \right)}=i \\\ & \Rightarrow \dfrac{3x+3i\left( x-2 \right)-ix+\left( x-2 \right)+6y-3i\left( 3y-1 \right)+2iy+\left( 3y-1 \right)}{\left( 9+1 \right)}=i \\\ & \Rightarrow 3x+3ix-6i-ix+x-2+6y-9iy+3i+2iy+3y-1=10i \\\ \end{aligned}$$ $$\Rightarrow 3x+3ix-6i-ix+x-2+6y-9iy+3i+2iy+3y-1-10i=0$$ Now we will separate out real and imaginary terms of the equation $$\begin{aligned} & \Rightarrow \left( 3x+x-2+6y+3y-1 \right)+\left( 3ix-6i-ix-9iy+3i+2iy-10i \right)=0 \\\ & \Rightarrow \left( 3x+x-2+6y+3y-1 \right)+i\left( 3x-6-x-9y+3+2y-10 \right)=0 \\\ & \Rightarrow \left( 4x+9y-3 \right)+i\left( 2x-7y-13 \right)=0 \\\ \end{aligned}$$ As we know that $$0$$ is a complex number which we can expressed as $$0=0+i0$$ $$\Rightarrow \left( 4x+9y-3 \right)+i\left( 2x-7y-13 \right)=0+i0$$ On comparing real part and imaginary part, we get $$\begin{aligned} & \Rightarrow \left( 4x+9y-3 \right)=0\cdots \cdots \cdots \left( i \right) \\\ & \Rightarrow \left( 2x-7y-13 \right)=0\cdots \cdots \cdots \left( ii \right) \\\ \end{aligned}$$ On solving both equations, we get $$\Rightarrow x=3\text{ and }y=-1$$ Hence, the real values of $$x$$ and $$y$$ are $$x=3\text{ and }y=-1$$ But we have given that the real values of $$x$$ and $$y$$ are represented by $$\lambda $$ and $$\mu $$ respectively $$\Rightarrow \lambda =3\text{ and }\mu =-1$$ Now we have to find the value of $$\lambda -\mu $$, so let us consider $$\begin{aligned} & \Rightarrow \lambda -\mu =3-\left( -1 \right) \\\ & \Rightarrow \lambda -\mu =3+1 \\\ & \Rightarrow \lambda -\mu =4 \\\ \end{aligned}$$ Hence, if the real values of x and y, for $$\dfrac{\left( 1+i \right)x-2i}{3+i}+\dfrac{\left( 2-3i \right)y+i}{3-i}=i$$ are $$\lambda $$ and $$\mu $$ respectively, then $$\lambda -\mu $$ is $$4$$ **Note:** In this type of question students have to remember that $$0$$ can be considered as a complex number which can be expressed in the form $$0=0+i0$$. Students have to note the values of $$i$$ and $${{i}^{2}}$$. Also students have to remember to find the value of $$\lambda -\mu $$ after finding the real values of $$x$$ and $$y$$.