Question:
If $z=a+i b$ lies in third quadrant, then $\frac{\bar{z}}{z}$ also lies in third quadrant if
(a) $a>b>0$
(b) $a (c) $b (d) $b>a>0$
Solution:
Since, $z=a+i b$ lies in third quadrant.
$\Rightarrow a<0$ and $b<0$
Now,
$\frac{\bar{z}}{z}=\frac{\overline{a+i b}}{a+i b}$
$=\frac{a-i b}{a+i b}$
$=\frac{a-i b}{a+i b} \times \frac{a-i b}{a-i b}$
$=\frac{a^{2}+i^{2} b^{2}-2 a b i}{a^{2}-i^{2} b^{2}}$
$=\frac{a^{2}-b^{2}-2 a b i}{a^{2}+b^{2}}$
Since, $\frac{\bar{z}}{z}$ also lies in third quadrant.
$\Rightarrow a^{2}-b^{2}<0$
$\Rightarrow(a-b)(a+b)<0$
$\Rightarrow a-b>0$ and $a+b<0$
$\Rightarrow a>b \quad \ldots(2)$
From (1) and (2),
$b Hence, the correct option is (c).