How can I find the complex numbers satisfying this condition? For a given complex number $a$ with $|a|\ge1,$ I want to find the all complex numbers on the unit circle such that $$\dfrac{z}{(a-z\bar a)^2}\in\mathbb{R}$$ and satisfying the condition $$\dfrac{z}{(a-z\bar a)^2}\le-\dfrac{1}{4}.$$ 
I express $z$ as $x+iy,$ but resulting expression is not easy to solve.
How can we  solve these kind of problems?
 A: Write $a = re^{i\varphi}$ with $r > 0$ and $\varphi \in \mathbb{R}$. An easy computation shows that
$$f(z) = \frac{z}{(a-\overline{a}\cdot z)^2} = \frac{1}{r^2}\frac{e^{-2i\varphi}z}{(1-e^{-2i\varphi}z)^2} = \frac{1}{r^2} K(e^{-2i\varphi}z),$$
where
$$K(w) = \frac{w}{(1-w)^2}$$
is the Koebe function. It is relatively well-known (and if you don't know it, it's a good exercise to prove it) that the Koebe function is a Schlicht function minimising the disk around $0$ contained in its image, and that
$$K\colon \mathbb{D} \to \mathbb{C} \setminus \left(-\infty, -\tfrac{1}{4}\right]$$
is biholomorphic.
So all points on the unit circle except for $1$, where $K$ has a pole, are mapped to $\left(-\infty,-\frac{1}{4}\right]$ by $K$, and if $r = \lvert a\rvert = 1$, then the same is true of $f$ [up to rotation, the pole is $e^{2i\varphi}$]. If $r > 1$, we want to find the points on the unit circle that $K$ maps to $\left(-\infty,-\frac{r^2}{4}\right]$ (and then rotate by $e^{2i\varphi}$). So let's invert $K$ (suppose $w\neq 0$):
\begin{align}
&& w &= \frac{z}{(1-z)^2}\\
&\iff& w(z^2-2z+1) &= z\\
&\iff& z^2 - 2\left(1 + \tfrac{1}{2w}\right)z + 1 &= 0\\
&\iff& \left(z - \left(1 + \tfrac{1}{2w}\right)\right)^2 &= \frac{4w+1}{4w^2}\\
&\iff& z &= \frac{2w+1\pm \sqrt{4w+1}}{2w}.
\end{align}
Note that the possible values of $z$ are reciprocals of each other, so if one of them lies on the unit circle, both do.
Now insert $w = -\frac{t}{4}$ for $t \geqslant r^2$ and obtain
$$K^{-1}\left(-\tfrac{t}{4}\right) = \left\{ \frac{1-t/2 \pm \sqrt{1-t}}{-t/2}\right\} = \left\{ 1-\tfrac{2}{t} \pm \frac{2i}{t}\sqrt{t-1}\right\}.$$
The two values describe two arcs on the unit circle, one in the upper, and one in the lower half-plane, approaching $1$ for $t\to\infty$, and $-1$ for $t\to 1$.
Inserting specifically $t = r^2$ yields the end points of the arc,
$$1 - \frac{2}{r^2} \pm \frac{2i}{r^2}\sqrt{r^2-1}.$$
If we  want to find the angle, which seems most convenient, we obtain
$$\alpha_{r^2} = \operatorname{arccot} \frac{r^2-2}{\sqrt{r^2-1}},$$
and
$$K^{-1}\left(\left(-\infty, -\tfrac{r^2}{4}\right]\right) = \left\{ z \in \partial \mathbb{D} : 0 < \lvert \arg z\rvert \leqslant \operatorname{arccot} \frac{r^2-2}{\sqrt{r^2-1}}\right\}.$$
Hence
$$f^{-1}\left(\left(-\infty,-\tfrac{1}{4}\right]\right) =  \left\{ z \in \partial \mathbb{D} : 0 < \lvert \arg z - 2\arg a\rvert \leqslant \operatorname{arccot} \frac{\lvert a\rvert^2-2}{\sqrt{\lvert a\rvert^2-1}}\right\}.$$
A: $$\dfrac{z}{(a-z\bar a)^2}\le-\dfrac{1}{4}$$
Let $a=re^{i\varphi}$ where $r\ge1$ with the usual meanings. Then 
$$\frac{z}{(a-\overline{a}\cdot z)^2} = \frac{1}{r^2}\frac{e^{-2i\varphi}z}{(1-e^{-2i\varphi}z)^2} $$
Note that since we are concern about $z$ with $|z|=1,$ always $e^{-2i\varphi}z$ is lie on the unit circle.
Let $$e^{-2i\varphi}z=e^{i\psi}=\cos\psi+i\sin\psi $$ with $\psi\not= 0,$ because $z\not=\frac{a}{\bar a}.$ Then 
$$\frac{1}{r^2}\frac{e^{-2i\varphi}z}{(1-e^{-2i\varphi}z)^2}=\frac{1}{r^2}\frac{e^{i\psi}}{(1-e^{i\psi})^2}.$$
(Using D'Moivers theorem) We can show that 
$$\dfrac{z}{(a-z\bar a)^2}=-\dfrac{1}{4r^2\sin^2\frac{\psi}{2}}$$ Then we have to find bounds for $\psi$ interms of $r,$ using the condition $$-\dfrac{1}{4r^2\sin^2\frac{\psi}{2}}\le-\dfrac{1}{4}\\r^2\sin^2\frac{\psi}{2}-1\le0\\-\dfrac{1}{r}\le\sin\frac{\psi}{2}\le\dfrac{1}{r}$$
Finally this gives us $$|\psi|\le2\operatorname{arccosec} r $$
Hence we can obtain, our $z$ should lie on the arc $$\{ z \in \partial \mathbb{D} :0<|\arg z-2\arg a|\le2\operatorname{arccosec}|a|\}$$
