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

Hint: 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{align}
  & z-\overline{z}=a+ib-\left( a-ib \right) \\
 & z-\overline{z}=a+ib-a+ib \\
 & z-\overline{z}=2ib \\
\end{align} $ and $ \begin{align}
  & z+\overline{z}=a+ib+\left( a-ib \right) \\
 & z+\overline{z}=a+ib+a-ib \\
 & z+\overline{z}=2a \\
\end{align} $
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{align}
  & \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{align} $
As we know that $ {{i}^{2}}=-1 $ so replace it in above equation, so we will get;
 $ \begin{align}
  & 6a+2ia-4ib-2{{i}^{2}}b+14i=0 \\
 & 6a+2ia-4ib+2b+14i=0 \\
\end{align} $
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{align}
  & 6a+2b=0 \\
 & a=\dfrac{-2b}{6} \\
 & a=\dfrac{-b}{3} \\
 & b=-3a \\
\end{align} $
Now put the value of ‘b’ in terms of ‘a’ we got from above equation in imaginary part, so we will have;
 $ \begin{align}
  & 14+2a-4b=0 \\
 & 14+2a-4\left( -3a \right)=0 \\
 & a=-1 \\
\end{align} $
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{align}
  & 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{align} $
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} $ .