I am completely stuck in the following problem:
let f be holomorphic function on the unit disk $\{z:|z|\lt 1\}$ in the complex plane.which of the following is/are necessarily true?
(a) if for each positive integer $n$ we have $f(\frac{1}{n})=\frac{1}{n^2}$,then $f(z)=z^2$ on the unit disc.
(b) if for each positive integer $n$ we have $f(1-\frac{1}{n})=(1-\frac{1}{n})^2$,then $f(z)=z^2$ on the unit disc.
(c) $f$ cannot satisfy $f(\frac{1}{n})=\frac{(-1)^n}{n}$ for each positive integer $n$
(d) $f$ cannot satisfy $f(\frac{1}{n})=\frac{1}{n+1}$ for each positive integer n
what I thought in this problem is the following:
in the first part,by identity theorem we can say that $f(z)=z^2,\forall$z in the unit disc.
but,in the second part ,if we choose $g(z)=z^2$ then,$f(1-\frac{1}{n})=g(1-\frac{1}{n})$ but,the limit point ,i.e, $1$ is not in the unit disc.so,in this case,we cannot use identity theorem to conclude whether $f(z)=z^2$ on the unit disk or not .so,what procedure can we follow in this case???also,explain the options (c) and (d).
please answer in detail..
thanks a lot…
Best Answer
For (D) , Consider , $\displaystyle g(z)=f(z)-\frac{z}{z+1}$. Then , $g(1/n)=0$and $\{1/n\}$ has a limit point $0\in \mathbb D$. So, $\displaystyle f(z)=\frac{z}{z+1}$ , which is analytic in $\{z:|z|<1\}$. So (D) is FALSE.
For (C) , put $\frac{1}{n}=z$. Then , $f(z)=z(-1)^{1/z}$ , which is NOT analytic at $z=0$. So (C) is TRUE.
For (B) , consider the function $\displaystyle f(z)=z^2+\sin\left(\frac{\pi}{1-z}\right)$ , which is analytic in $\{|z|<1\}$ satisfying your condition.