Is the function f(z)=dfrac11-z uniformly continuous on the…

Is the function $\ f(z)=\dfrac1{1-z}\ $ uniformly continuous on the disk $|z|<1$?

Step 1 · Definition being tested

$f$ is uniformly continuous on $D$ if for every $\varepsilon>0$ there is $\delta>0$ (independent of the points) such that $|z_1-z_2|<\delta\implies|f(z_1)-f(z_2)|<\varepsilon$ for all $z_1,z_2\in D$.

Step 2 · A general estimate near $z=1$

$$ f(z_1)-f(z_2)=\frac1{1-z_1}-\frac1{1-z_2}=\frac{z_1-z_2}{(1-z_1)(1-z_2)} $$

If both $z_1,z_2$ are close to the boundary point $z=1$, the denominator $(1-z_1)(1-z_2)$ can be made arbitrarily small independently of $|z_1-z_2|$, so the ratio can be made arbitrarily large even while $|z_1-z_2|\to0$.

Step 3 · Explicit counterexample sequences

Let $z_n=1-\dfrac1n$ and $w_n=1-\dfrac1n-\dfrac1{n^2}$. Then $$ |z_n-w_n|=\frac1{n^2}\to0 $$ while $$ f(z_n)=n,\qquad f(w_n)=\frac1{\tfrac1n+\tfrac1{n^2}}=\frac{n^2}{n+1} $$ $$ |f(z_n)-f(w_n)|=\left|n-\frac{n^2}{n+1}\right|=\frac n{n+1}\to1 $$

So for $\varepsilon=\tfrac12$, no matter how small $\delta>0$ is chosen, taking $n$ large enough makes $|z_n-w_n|<\delta$ while $|f(z_n)-f(w_n)|\to1>\varepsilon$ — uniform continuity fails.

Geometric reason

$f$ has a simple pole exactly at the boundary point $z=1$; $f$ is continuous (indeed holomorphic) at every point of the open disk, but it cannot be extended continuously to the closed disk $|z|\le1$, and a continuous function on a bounded open set is uniformly continuous precisely when such a continuous extension to the closure exists.