Question

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

• A. $5$
• B. $8$
• C. $10$
• D. $40$

Answer 1

Answer: (C)

$10$

Answer 2

Let $z=x+iy,$ then $\bar{z}=x-iy$
$\therefore$ $z+\bar{z}=2x$ and $z-\bar{z}=2iy$
Given, $(3+i)(z+\bar{z})-(2+i)(z-\bar{z})+14i=0$
$\Rightarrow$ $(3+i)2x-(2+i)2iy+14i=0$
$\Rightarrow$ $6x+2ix-4yi+2y+14i=0+0i$
On comparing real and imaginary part, we get
$6x+2y=0$ and $2x-4y+14=0$
On solving, we get
$x=-1,\,y=3$
$\therefore$ $z\bar{z}=|z{{|}^{2}}={{(\sqrt{{{(-1)}^{2}}+{{(3)}^{2}}})}^{2}}=10$