Existence of analytic function satisfying given conditions

Question

Let $\left({z_n}\right)$ be a sequence of distinct points in $D\left(0,1\right)$=$\{{z \in \mathbb{C} :\vert{z}\vert \lt{1}} \}$ with $\lim_{n\to\mathbb{\infty}} {z_n}=0$. Consider the following statements P and Q.

P: There exists a unique analytic function $f$ on $D\left(0,1\right)$ such that $f\left({z_n}\right)=\sin\left({z_n}\right)$ for all $n $.

Q: There exists an analytic function $f$ on $D\left(0,1\right)$ such that $f\left({z_n}\right)=0$, if $n$ is even and $f\left({z_n}\right)=1$, if $n$ is odd.

Which of the above statements hold true?

For P : I know that if two functions agree on convergent sequence in domain then they agree on whole domain.

Let ${z_n}=\frac{1}{n+1}$ then $\lim_{x\to \infty}{z_n}=0$ and $\lim_{x\to \infty} f(z_n)=\sin(z_n)=0$ so what I can conclude here?

For Q : I think this function is not continuous hence not analytic. But I am not getting how to prove it is not continuous?

Answer 1

Hint 1: The analytic functions $f$ and $\sin(z)$ coincide along the given converging sequence $(z_n)_n$.

Hint 2. If $f$ is continuous at $0$ then for any sequence $(z_n)_n$ such that $z_n\to 0$ we have that $f(z_n)\to f(0)$. Consider the sequences $(z_{2n})_n$ and $(z_{2n+1})_n$ which converge to $0$. What are the limits $$\lim_{n\to\infty}f(z_{2n})=?\qquad,\qquad \lim_{n\to\infty}f(z_{2n+1})=?$$ What may we conclude about the value of $f$ at $0$?

Answer 2

P is true take $f(z) =\sin(z)$ and $z_n -\frac{1}{2^n}$

Suppose that Q is true ans set $a_n=z_{2n}$ and $b_n =z_{2n+1}$ Then, $$a_n,b_n\to 0$$ and by continuity,

$$ 0=\lim_{n\to\infty} f(a_n) = f(\lim_{n\to\infty} a_n) = f(0) =f(\lim_{n\to\infty} b_n) =\lim_{n\to\infty} f(b_n) =1$$ contradiction.