Let, $D=\{z\in \mathbb C:|z|<1\}$. Then there exists a non-constant analytic function$f$ on $D$ such that for all $n=2,3,4,...$
(a) $f\left(\frac{i}{n}\right)=0$.
(b) $f\left(\frac{1}{n}\right)=0$.
(c) $f\left(1-\frac{1}{n}\right)=0$.
(d) $f\left(\frac{1}{2}-\frac{1}{n}\right)=0$.
Which is(/are) correct(/s).
I tried with Identity theorem, but I have some confusion.
From Identity theorem, $S=$ set of all zeros of $f$ has a limit point in $D$ for the options (a) , (b) & (d). So these functions are identically zero. As the function is non-constant so these are NOT possible. So option (c) is only correct.
Am I right? OR wrong?
Here is a hint for c. You want $f$ to not be analytic at $z=1$ therefore you may want to try and play with $z-1$ in a denominator. Now for the points $z=1-\frac{1}{n}$ we have $z-1=-\frac{1}{n}$ so the denominator conveniently disappears and lastly you want $f(1-\frac{1}{n})=0$ for all integers so try to think about periodic functions.