Question

If $z=x+iy$ lies in $3rd$ quadrant, then $\frac{\overline{z}}{z}$ also lies in the $3rd$ quadrant if

• A. $y>0>x$
• B. $y<x<0$
• C. $x<y<0$
• D. $x>0>y$

Answer 1

The correct option is C $x<y<0$

Since $z=x+iy$ lies in the third quadrant

$\therefore x<0,y<0$

Again $\overline{z}=x-iy$

$\therefore \frac{\overline{z}}{z}=\frac{x-iy}{x+iy}=\frac{(x-iy)(x-iy)}{x^2+y^2}=\frac{x^2-y^2-2ixy}{x^2+y^2}$

$=\frac{x^2-y^2}{x^2+y^2}-\frac{2xy}{x^2+y^2}i=A+iB$

where $A=\frac{x^2-y^2}{x^2+y^2}$ and $B=-\frac{2xy}{x^2+y^2}$

Since $x<0,y<0$

$\therefore -\frac{2xy}{x^2+y^2}<0\Rightarrow B<0$

Now, $\frac{\overline{z}}{z}$ lies in the $3rd$ quadrant if $a<0$

i.e., if $x^2-y^2<0\Rightarrow x^2<y^2$

i.e., if $x<y<0$