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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.

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    What is your definition of "convex" for a complex function?2011-12-14
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    Now, what is $t$ that you've added to the definition of $k$?2011-12-14
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    @HenningMakholm: already put it in my details question2011-12-14
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    @ThomasAndrews:I just at the beginning stage to understand this, that is all they have been wrote in that journal.2011-12-14

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