# Lower and upper bounds for fractional linear transformation

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

$$k(z)=\frac{z}{1-tz} = -\frac{1}{t} + \frac{1}{t(1-tz)}$$
Now test with some points in $\mathbb{C}$ such as $\infty,0,1,\frac{1}{t}$ etc.
Select circle and enter the variables at the bottom with Enter new values button.