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If we have $k(z)=\frac{z}{1-tz}$ which is convex in unit disk, then $k(\bar{z})=\overline{k(z)}$, $k(z)$ maps real axis to real axis where $|z|\leq{r}$, $t\in\mathbb{R}$. What is the upper and lower bounds of $\Re {k(z)}$?

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If $r\ge1/t$ then there can't be any upper bound since $k\to\infty$ as $z\to1/t$ –  Gerry Myerson Dec 2 '11 at 2:59
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1 Answer

I would suggest writing the LFT in the following form:

$$ k(z)=\frac{z}{1-tz} = -\frac{1}{t} + \frac{1}{t(1-tz)} $$

Now test with some points in $\mathbb{C}$ such as $\infty,0,1,\frac{1}{t}$ etc.

For some geometric aid, to see Gerry Myerson's comment try this applet. Also you can find your answer by trying out.

Select circle and enter the variables at the bottom with Enter new values button.

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