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 ?
 A: 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.
A: Hint
Write $\dfrac{\bar{z}}{z} = \dfrac{\bar{z}^2}{|z|^2}$ which we want in the third quadrant. Since the denominator is real, this means arg($\bar{z}$) must be between $90$ and $135$ degrees (for $\bar{z}^2$ to be between $180$ and $270$ degrees). This translates to condition $y \lt x \lt 0$, so you are correct.
A: The approach in the OP is solid.  I thought it might be instructive to present a way forward using polar coordinates.
First, let $z=re^{i\theta}$. Next, we see that $\bar z = re^{-i\theta}$ and therefore
$$\frac{\bar z}z=e^{-i2\theta}$$
In order for $\bar z/z$ to be in Quadrant 3, we must have $-2\theta \in (\pi+2n\pi,3\pi/2+2n\pi)$ for some integer $n$.  This implies that $\theta \in (-3\pi/4-n\pi,-\pi/2-n\pi)$.  
Given that $z$ is in the third quadrant, $n$ must be even.  Taking $n=0$, we have $\theta \in (-3\pi/4,-\pi/2)$, which occurs when $y<x< 0$.
A: If $180^{\circ}<\theta<270^{\circ}$ then $-540^{\circ}<-2\theta<-360^{\circ}$,
which gives $-540^{\circ}<-2\theta<-450^{\circ}$ or $225^{\circ}<\theta<270^{\circ}$.
$\sin$ decreases and $\cos$ increases,  which gives $-1<\sin\theta<-\frac{1}{\sqrt2}$ and $-\frac{1}{\sqrt2}<\cos\theta<0$.
Thus, $y<x<0$.
A: I get $-\frac \pi2\gt \theta \gt \frac{-3\pi}4$, which is the same as $z=x+iy $ in the 3rd quadrant and $y\lt x \lt 0$...
My method :  use polar coordinates,  $z=re^{i\theta}$, $\pi \lt \theta \lt\frac {3\pi }2$.  Then  $\frac 
 {\bar z} z=\frac {re^{-i\theta}}{re^{i\theta}}=e^{-2i\theta }$.  So $\pi\lt -2\theta \lt \frac{3\pi}2$.
A: You cannot, since $x<y<0$ is wrong.
Geometrically, what it tells you that $z = x + iy$ when written in form $z = re^{i\varphi}$ satisfies $\varphi\in(\pi,\frac{5\pi}4)$. Here's a graph to support the claim:

Note that $\frac{\bar z}z = \frac{re^{-i\varphi}}{re^{i\varphi}} = e^{-2i\varphi}$. Now, $\pi<\varphi<\frac{5\pi}4$ implies $-\frac{5\pi}2<-2\varphi<-2\pi$, and thus $\frac{\bar z}z$ is in fourth quadrant, not third.
Thus, your solution is correct, i.e. $y < x < 0$, and you can prove it similarly to what I've done here.
