Question

If $\left( 3+i \right)\left( \overline{z}+z \right)-\left( 2+i \right)\left( z-\overline{z} \right)+14i=0,$ then $\overline{z}z$ is equal to
A. 5
B. 8
C. 10
D. 40

Answer 1

We will solve the complex number equation as a normal equation, by using some of the properties of complex numbers, that are ‘z’ is defined as a $a+ib$ , and $\overline{z}$ is $a-ib$ .

Complete step by step answer:
Moving ahead with the question in step wise manner;
As we know that the ‘z’ is represent a complex number which is a combination of real value and imaginary value i.e. $z=a+ib$ in which ‘a’ is the real value and ‘b’ is the imaginary value. And as we also know that if $z=a+ib$ then $\overline{z}=a-ib$ .
So we can say that $z-\overline{z}$ and $z+\overline{z}$ is
$\begin{aligned} & z-\overline{z}=a+ib-\left( a-ib \right) \\\ & z-\overline{z}=a+ib-a+ib \\\ & z-\overline{z}=2ib \\\ \end{aligned}$ and $\begin{aligned} & z+\overline{z}=a+ib+\left( a-ib \right) \\\ & z+\overline{z}=a+ib+a-ib \\\ & z+\overline{z}=2a \\\ \end{aligned}$
So we can say that $z-\overline{z}$ and $z+\overline{z}$ will be always $2ib$ and $2a$
So let us solve the above equation as assuming it as simple equation, so we will get;
$\begin{aligned} & \left( 3+i \right)\left( \overline{z}+z \right)-\left( 2+i \right)\left( z-\overline{z} \right)+14i=0 \\\ & \left( 3+i \right)2a-\left( 2+i \right)2ib+14i=0 \\\ & 6a+2ia-4ib-2{{i}^{2}}b+14i=0 \\\ \end{aligned}$
As we know that ${{i}^{2}}=-1$ so replace it in above equation, so we will get;
$\begin{aligned} & 6a+2ia-4ib-2{{i}^{2}}b+14i=0 \\\ & 6a+2ia-4ib+2b+14i=0 \\\ \end{aligned}$
On comparing the real and imaginary part we will get;
$6a+2b+i\left( 14+2a-4b \right)=0$
So real part we have;
$\begin{aligned} & 6a+2b=0 \\\ & a=\dfrac{-2b}{6} \\\ & a=\dfrac{-b}{3} \\\ & b=-3a \\\ \end{aligned}$
Now put the value of ‘b’ in terms of ‘a’ we got from above equation in imaginary part, so we will have;
$\begin{aligned} & 14+2a-4b=0 \\\ & 14+2a-4\left( -3a \right)=0 \\\ & a=-1 \\\ \end{aligned}$
So here we got ‘a’ equal to -1 which is the real value of ‘z’. Now put the value of ‘a’ in $a=\dfrac{-b}{3}$ to get the value of ‘b’. So we will get ‘b’ equal to $3$ .
So we will get the value $z$ which is equal to $a+ib$ , and as from the above solution we get the value of ‘a’ equal to 7 and value of ‘b’ equal to -21. So we got $z=-1+3i$ . Similarly$\overline{z}=-1-3i$.
So we need to find out the value of $z\overline{z}$ . So just put the value of ‘z’ and $'\overline{z}'$ so we will get;
$\begin{aligned} & z\overline{z}=\left( -1+3i \right)\left( -1-3i \right) \\\ & z\overline{z}={{\left( -1 \right)}^{2}}-{{\left( 3i \right)}^{2}} \\\ & z\overline{z}=1-\left( -9 \right) \\\ & z\overline{z}=10 \\\ \end{aligned}$
So we got $z\overline{z}=10$ .

So, the correct answer is “Option C”.

Note: We will always get the value of $z-\overline{z}$ and $z+\overline{z}$ constant which will be $2ib$ and $2a$ respectively, in which ‘a’ is the real value of complex number ‘z’ and b is the imaginary value of complex number ‘z’ and $i$ is the iota denoted as $\sqrt{-1}$ .