<s>[INST] Under what conditions $z=x+iy$ in $3^{rd}$ quadrant implies $\frac{\bar{z}}{z}$ in $3^{rd}$ quadrant? 
If $z=x+iy$ is in the $3^{rd}$ quadrant, then $\frac{\bar{z}}{z}$ also lies in the $3^{rd}$ quadrant if ?

How do I obtain the solution which is given as $x<y<0$.
My Attempt:
$x=|z|\cos\theta<0$ & $y=|z|\sin\theta<0$
Since $\frac{\bar{z}}{z}=\frac{|\bar{z}|}{|z|}[\cos(-2\theta)+i\sin(-2\theta)]$ is in the $3^{rd}$ quadrant,
$$
\cos(-2\theta)<0 \text{ & } \sin(-2\theta)<0\implies \cos^2\theta-\sin^2\theta<0\text{ & }\sin\theta\cos\theta<0
$$
From $\sin\theta\cos\theta<0$ we only take $x<0$ & $y<0$ since it is already stated. Now,
$$
\cos^2\theta-\sin^2\theta<0\implies\cos^2\theta<\sin^2\theta\implies|\cos\theta|<|\sin\theta|\implies|x|<|y|
$$
Since both $x,y<0$, I think $|x|=-x$ & $|y|=-y$
$$
|x|<|y|\implies-x<-y\implies x>y
$$
So the solution I obtain is $y<x<0$.
Is this the right way to approach the inequalities of this kind ?. What am I doing wrong ?
 [/INST] Note that a complex number $a+ib$ is in the third quadrant (not including axes) iff $a,b<0$. Therefore we can assume that $x,y<0$ in order to have $z$ in the third quadrant, and we want to find conditions so that 
$\frac{\bar{z}}{z}$ is in the third quadrant. Now
$$
\frac{\bar{z}}{z} = \frac{x-iy}{x+iy} = \frac{x^2-y^2-ixy}{x^2+y^2}
= \frac{x^2-y^2}{x^2+y^2} +i\frac{-xy}{x^2+y^2},
$$
so we need to have $x^2-y^2<0$ and $-xy<0$. The latter is true because $x,y<0$, and the former is equivalent to $y<x$ since we know $x$ and $y$ are negative.
We conclude that $y<x<0$ are the desired conditions.
The conditions $x<y<0$, which appear in the question, don't work. Indeed, consider $z=-2-i$. Then $z$ is in the third quadrant, but
$$
\frac{\bar{z}}{z} = \frac{-2+i}{-2-i} = \frac{3-4i}{4+1} = \frac{3}{5}-\frac{4}{5}i
$$
is in the fourth quadrant.
</s>