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If $z=x+i y$ lies in the third quadrant, then $\frac{\bar{z}}{z}$ also lies in the third quadrant, if
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Verified Answer
The correct answer is:
$x < y < 0$
$x < y < 0$
$\mathrm{As}, \mathrm{z}=\mathrm{x}+$ iy lies in the third quadrant. So, $x < 0$ and $y < 0$.
$$
\begin{aligned}
&\because \quad \frac{\bar{z}}{z}=\frac{x-i y}{x+i y}=\frac{(x-i y)(x-i y)}{(x+i y)(x-i y)}=\frac{x^2-y^2-2 i x y}{x^2+y^2} \\
&=\frac{\bar{z}}{z}=\frac{x^2-y^2}{x^2+y^2}-\frac{2 i x y}{x^2+y^2}
\end{aligned}
$$
As, $\frac{\bar{z}}{z}$ also lies in the third quadrant.
Therefore, $\frac{x^2-y^2}{x^2+y^2} < 0$ and $\frac{-2 x y}{x^2+y^2} < 0$ $\Rightarrow x^2-y^2 < 0$ and $-2 x y < 0\left[\because x^2+y^2>0\right]$ $\Rightarrow x^2 < y^2$ and $x y>0$. Hence, $x < y < 0$
$$
\begin{aligned}
&\because \quad \frac{\bar{z}}{z}=\frac{x-i y}{x+i y}=\frac{(x-i y)(x-i y)}{(x+i y)(x-i y)}=\frac{x^2-y^2-2 i x y}{x^2+y^2} \\
&=\frac{\bar{z}}{z}=\frac{x^2-y^2}{x^2+y^2}-\frac{2 i x y}{x^2+y^2}
\end{aligned}
$$
As, $\frac{\bar{z}}{z}$ also lies in the third quadrant.
Therefore, $\frac{x^2-y^2}{x^2+y^2} < 0$ and $\frac{-2 x y}{x^2+y^2} < 0$ $\Rightarrow x^2-y^2 < 0$ and $-2 x y < 0\left[\because x^2+y^2>0\right]$ $\Rightarrow x^2 < y^2$ and $x y>0$. Hence, $x < y < 0$
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