If z=a+ib lies in third quadrant, then

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).