Show that if $f$ is analytic in the unit disc then an integer $n$ such that $f(1/n)$ does not equal $1/(n+1)$

Question

This is a variant of question Show that if f is analytic in $|z|\leq 1$, there must be some positive integer n such that $f(\frac{1}{n})\neq \frac{1}{n+1}$.

(i). Show that if $f$ is analytic in the unit disc $D=\{z\in C : |z|<1\}$, then there exists an integer $n$ such that $f(1/n)\neq1/(n+1)$
(ii) Does there exist an analytic function $g$ in the unit disc $D=\{z\in C : |z|<1\}$ such that $g(1/n)=g(-1/n)=1/n^2$ for all positive integers $n$?
(iii) Does there exist an analytic function $g$ in the unit disc $D=\{z\in C : |z|<1\}$ such that $h(1/n)=h(-1/n)=1/n^3$ for all positive integers $n$?

Thoughts thus far: For (i) the linked problem seems to be the most concise way to show this, but we are noy concerned with the point $z = -1$ as it was in the linked problem so the.same method cannot be used as that in the linked problem For (ii) and (iii), isn't the answer no because there is a singularity at the origin. However, perhaps I am missing some lemma not shown in class that shows an analytic function may have a removable discontinuity (but I doubt this is the case).

Thank you in advance for any help that you may provide.

Answer 1

More or less obvious candidate analytic functions for your conditions are

• $f(z)=\frac z{1+z}$ with pole at $z=-1$, i.e. luckkily outside $D$.
• $g(z)=z^2$

For the third, we must have $h(z)=z^3$ because that holds for the sequence $z_n=\frac1n$ with $z_n\to 0$. However, this contradicts $h(-\frac1n)=\frac1{n^3}$. Hence there is no such function.