Question

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

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

Answer 1

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

Solution -

$z=x+iy$ lines in third quadrant

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

$\frac{\overline{z}}{z}=\frac{x^2-y^2}{x^2+y^2}-\frac{2ixy}{x^2+y^2}$

For it to be in third quadrant

$y<x<0$