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 $x&lt;y&lt;0$.

My Attempt:

$x=|z|\cos\theta&lt;0$ &amp; $y=|z|\sin\theta&lt;0$

Since $\frac{\bar{z}}{z}=\frac{|\bar{z}|}{|z|}[\cos(-2\theta)+i\sin(-2\theta)]$ is in the $3^{rd}$ quadrant,
$$
\cos(-2\theta)&lt;0 \text{ &amp; } \sin(-2\theta)&lt;0\implies \cos^2\theta-\sin^2\theta&lt;0\text{ &amp; }\sin\theta\cos\theta&lt;0
$$
From $\sin\theta\cos\theta&lt;0$ we only take $x&lt;0$ &amp; $y&lt;0$ since it is already stated. Now,
$$
\cos^2\theta-\sin^2\theta&lt;0\implies\cos^2\theta&lt;\sin^2\theta\implies|\cos\theta|&lt;|\sin\theta|\implies|x|&lt;|y|
$$
Since both $x,y&lt;0$, I think $|x|=-x$ &amp; $|y|=-y$
$$
|x|&lt;|y|\implies-x&lt;-y\implies x&gt;y
$$
So the solution I obtain is $y&lt;x&lt;0$.

Is this the right way to approach the inequalities of this kind ?. What am I doing wrong ?
Note that a complex number $a+ib$ is in the third quadrant (not including axes) iff $a,b&lt;0$. Therefore we can assume that $x,y&lt;0$ 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 $x^2-y^2&lt;0$ and $-xy&lt;0$. The latter is true because $x,y&lt;0$, and the former is equivalent to $y&lt;x$ since we know $x$ and $y$ are negative.

We conclude that $y&lt;x&lt;0$ are the desired conditions.

The conditions $x&lt;y&lt;0$, 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.
You cannot, since $x&lt;y&lt;0$ 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, $\pi&lt;\varphi&lt;\frac{5\pi}4$ implies $-\frac{5\pi}2&lt;-2\varphi&lt;-2\pi$, and thus $\frac{\bar z}z$ is in fourth quadrant, not third.
Thus, your solution is correct, i.e. $y &lt; x &lt; 0$, and you can prove it similarly to what I've done here.