Question

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

(a) $a>b>0$
(b) $a<b<0$
(c) $b<a<0$
(d) $b>a>0$

Answer 1

Since, $z=a+ib$ lies in third quadrant.
$\Rightarrow a<0\mathrm{and}b<0....\left(1\right)$

Now,
$$\begin{gathered} \frac{\overline{z}}{z}=\frac{\overline{a+ib}}{a+ib} \\ =\frac{a-ib}{a+ib} \\ =\frac{a-ib}{a+ib}\times \frac{a-ib}{a-ib} \\ =\frac{a^2+i^2{b}^2-2abi}{a^2-i^2{b}^2} \\ =\frac{a^2-b^2-2abi}{a^2+b^2} \end{gathered}$$

Since, $\frac{\overline{z}}{z}$ also lies in third quadrant.
$$\begin{gathered} \Rightarrow a^2-b^2<0 \\ \Rightarrow (a-b)(a+b)<0 \\ \Rightarrow a-b>0\mathrm{and}a+b<0 \\ \Rightarrow a>b....\left(2\right) \end{gathered}$$

From (1) and (2),
$b<a<0$

Hence, the correct option is (c).