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


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) $b>a>0$


Since, $z=a+i b$ lies in third quadrant.

$\Rightarrow a<0$ and $b<0$


$\frac{\bar{z}}{z}=\frac{\overline{a+i b}}{a+i b}$

$=\frac{a-i b}{a+i b}$

$=\frac{a-i b}{a+i b} \times \frac{a-i b}{a-i b}$

$=\frac{a^{2}+i^{2} b^{2}-2 a b i}{a^{2}-i^{2} b^{2}}$

$=\frac{a^{2}-b^{2}-2 a b i}{a^{2}+b^{2}}$

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

$\Rightarrow a^{2}-b^{2}<0$


$\Rightarrow a-b>0$ and $a+b<0$

$\Rightarrow a>b \quad \ldots(2)$

From (1) and (2),


Hence, the correct option is (c).

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