How to find the bound for this function? According to  journal entitled Certain subclass of starlike function by Gao and Zhou (2007), it was proven that $-\frac{r}{1+tr} \leq Re \{\frac{z}{1-tz} \}\leq \frac{r}{1-tr}$ where $|z|\leq r<1$ and $0 < t \leq 1$. It is possible to find the bound for $|\frac{z}{1-tz}|$ using that information?
 A: Edited. A first version of this question asked for a proof of the estimate given in Gao and Zhou's paper. Here is this proof:
I assume that $0<r<1$ and that $0< t\leq1$.
Since $$z\mapsto{\rm Re}{z\over1-tz}$$
is a harmonic function it takes its extrema on the boundary of the allowed domain, i.e., when $z=re^{i\phi}$. Therefore it is sufficient to consider the function
$$\eqalign{g(\phi)&:={\rm Re}{re^{i\phi}\over1-tre^{i\phi}}={r\cos\phi-tr^2\over 1-2rt\cos\phi+t^2r^2}\cr 
&={1-t^2 r^2\over 2t(1-2rt\cos\phi+t^2r^2)}-{1\over 2t}\qquad\qquad(0\leq\phi\leq\pi)\ .\cr}$$
On the right hand side the first denominator is increasing with increasing $\phi$. It follows that $g$ takes its maximum at $\phi=0$ and its minimum at $\phi=\pi$, and one easily obtains
$$g_\min=-{r\over 1+rt},\quad g_\max={r\over 1-rt}\ .\qquad\qquad\square$$
When only an estimate of
$$\left|{z\over 1-tz}\right|$$
is needed things are much simpler. Again the maximum of this  expression is taken when $|z|=r$. The points $tz$ then lie on a circle with radius $tr$, and among them the point nearest to $1$ is the point $tr$. It follows that
$$0\leq\left|{z\over 1-tz}\right|\leq{r\over 1-tr}\ .$$
