Question

If$(3+i)(z+\overline{z})-(2+i)(z-\overline{z})+14i=0$, then $z\overline{z}$is equal to

• A. $5$
• B. $8$
• C. $10$
• D. $40$

Answer 1

The correct option is C

$10$

Explanation for the correct option:

Step 1: Finding the values for $z+\overline{z},andz-\overline{z}$

It is given that $(3+i)(z+\overline{z})-(2+i)(z-\overline{z})+14i=0$

Let

$\begin{array}{rcl}z& =& x+iy\\ \overline{z}& =& x-iy\\ z+\overline{z}& =& (x+iy)+(x-iy)\\ & =& 2x\\ z-\overline{z}& =& (x+iy)-(x-iy)\\ & =& 2iy\end{array}$

Step 2: Finding the values of x and y

Substituting the values in given equations:

$\begin{array}{rcl}(3+i)2x-(2+i)2iy+14i& =& 0\end{array}$

$\Rightarrow$ $\begin{array}{rcl}6x+2ix-4iy+2y+14i& =& 0\end{array}$

$\Rightarrow$$\begin{array}{rcl}(6x+2y)+(2x-4y+14)i& =& 0\end{array}$

Comparing real part

$\begin{array}{rcl}6x+2y& =& 0\end{array}$

$\Rightarrow$ $\begin{array}{rcl}6x& =& -2y\end{array}$

$\Rightarrow$ $\begin{array}{rcl}y& =& \frac{6x}{-2}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}y& =& -3x\end{array}$

Comparing imaginary part

$\begin{array}{rcl}2x-4y+14& =& 0\end{array}$

Put $\begin{array}{rcl}y& =& -3x\end{array}$ in above equation

$\begin{array}{rcl}2x+12x+14& =& 0\end{array}$

$\Rightarrow$ $\begin{array}{rcl}14x& =& -14\end{array}$

$\Rightarrow$ $\begin{array}{rcl}x& =& -1\end{array}$

So, $\begin{array}{rcl}y& =& 3\end{array}$

Step 3: Finding the value for $z\overline{z}$

$\begin{array}{rcl}z& =& -1+3i\\ \overline{z}& =& -1-3i\end{array}$

$\begin{array}{rcl}\therefore z\overline{z}& =& (-1+3i)(-1-3i)\\ & =& 1+9\\ & =& \mathbf{10}\end{array}$

Hence, option (C) is the answer.