How to show it is convex? From a journal entitled Certain subclass of starlike functions by Gao and Zhou in 2007, they mentioned that  " since $ k(z)=\frac{z}{1-zt}$ is convex in open unit disk $E,z:|z|<1$, $k(\bar{z})= \bar{k(z)}$ and $k(z)$ maps real axis to real axis". How to show that it is convex in $E$ ?
The definition of convex function from Univalent function Volume 1 by A.W.Goodman: 
A set of domain, D in the plane is called convex if for every pair of point $w_{1}$ and $w_{2}$ in the interior of D, the line segment joining $w_{1}$ and $w_{2}$ is also in the interior of D. If a function $f(z)$ maps E onto a convex domain, then $f(z)$ is called a convex function.
Thank you.
 A: If your original definition, $k(z)=\frac{z}{1-z}$ is what you want, then $k$ is a Moebius transformation which send $1$ to $\infty$.  Any Moebius transformation is $1-1$ and onto the Riemann sphere, $\mathbb C \cup \{\infty\}$, and a Moebius transformation such that $k(1)=\infty$ has the property that any circle through $1$ gets mapped to a line, and the interiors and exteriors get mapped to half-planes on either side of that line. 
So $k$ maps the interior of the unit ball onto a half-plane, which is necessarily convex.
In general, any Moebius transformation $m(z)$ sends any ball either onto a half-plane, another ball, or the complement of the closure of the ball.   So for $m(z)$ to be convex on a ball, it is necessary and sufficient to prove that either $m(z_0)=\infty$ for some $z_0$ on the boundary of the ball, or $m(z)$ is bounded in the ball.
In the case $k_t(z)=\frac{z}{1-tz}$, this function is bounded on the unit ball if $|t|<1$ and has $z_0=\bar{t}$ with $h(z_0)=\infty$.  If $|t|>1$, then you can easily show that $k_t$ is unbounded on $E$.
So, $k_t$ is convex on $E$ if and only if $|t|\leq 1$.
For a general Moebius transformation, $m$, and ball, $B$, $m$ is convex on $B$ if and only if $m^{-1}(\infty)\notin B$.
