A Simple Question about Poles

Question

If $f$ be a meromorphic function on the complex plane and $(1-|z|)|f(z)|$ $\longrightarrow$ $0$ as $|z|$ $\longrightarrow$ 1 then show that $f$ does not have any pole on the unit circle. This looks very simple but I somehow have not able to prove it rigorously. Thanks for any help.

Answer 1

Assume a pole at $z=1$ so that $f(z)=\dfrac{g(z)}{(1-z)^n}$ with $g(1)$ finite and nonzero.

Then

$$\lim_{z\to1}(1-|z|)\left|\frac{g(z)}{(1-z)^n}\right|=g(1)\lim_{z\to1}\left|\frac{1-|z|}{(1-z)^n}\right|\ne 0.$$

For example,

$$\lim_{\epsilon\to0}\left|\frac{1-|1+\epsilon|}{\epsilon^n}\right|\ne 0.$$

Answer 2

Hint: If $|z_0|=1$ and $z\in \{rz_0: 0\le r \le1\},$ then $|z-z_0| = 1-|z|.$