Question

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

• A. a > b > 0
• B. a < b < 0
• C. b < a < 0
• D. b > a > 0

Answer 1

The correct option is B

a < b < 0

Given that, z = x + ib lies in third quadrant

x < 0 and y < 0

$\text{Now}\frac{\overline{z}}{z}=\frac{x-ib}{x+ib}=\frac{(x-ib)(x-ib)}{(x+ib)(x-ib)}=\frac{x^2-y^2-2ixb}{x^2+b^2}=\frac{\overline{z}}{z}=\frac{x^2-b^2}{x^2+b^2}-\frac{2ixb}{x^2+b^2}$

Since, $\frac{\overline{z}}{z}$ also lies in third quadrant.

$\therefore \frac{x^2-b^2}{x^2+b^2}<0\text{and}\frac{-2xb}{x^2+b^2}<0\therefore x^2-b^2<0\text{and}-2xb<0\text{So,}x<b<0$