Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
What is your definition of "convex" for a complex function? – Henning Makholm Dec 14 '11 at 15:21
Now, what is $t$ that you've added to the definition of $k$? – Thomas Andrews Dec 14 '11 at 15:38
@HenningMakholm: already put it in my details question – DRN Dec 14 '11 at 15:50
@ThomasAndrews:I just at the beginning stage to understand this, that is all they have been wrote in that journal. – DRN Dec 14 '11 at 15:56

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 $m^{-1}(\infty)\notin B$.

share|cite|improve this answer
yes,it is true that $|t|\leq 1$. I am not familiar with ball, $B$ and unit ball, is it same with disk or unit disk? – DRN Dec 14 '11 at 16:06
Yes, ball is just something we use in higher dimensions. Actually, a lot of places where I wrote "unit ball" I meant "ball" more generally, and everywhere I wrote "ball" you can just read it as "disk." – Thomas Andrews Dec 14 '11 at 16:10
Depending on what you definition of disk is, it might differ from the a disk in that a ball does not contain the boundary. So a ball is always a set of the form: $B_{z_0,r}=\{z\in\mathbb C: |z-z_0|<r\}$ – Thomas Andrews Dec 14 '11 at 16:16
Really good information. I am clear with your answers. Thank you so much. – DRN Dec 14 '11 at 16:20

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.