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Let $f$, $g$ be analytic function defined on $A\cup D$ where $A = \{z \in \mathbb{C}: \frac{1}{2}<|z|<1\}$ and $D = \{z \in \mathbb{C}: |z-2|<1\}$ Which of the following statements are true?

  1. If $f(z) g(z) =0$ for all $z \in A\cup D$, then either $f(z)=0$ for all $z \in A$ or $g(z) =0$ for all $z \in A$.
  2. If $f(z) g(z) =0$ for all $z \in D$, then either $f(z)=0$ for all $z \in D$ or $g(z) =0$ for all $z \in D$.
  3. If $f(z) g(z) =0$ for all $z \in A$, then either $f(z)=0$ for all $z \in A$ or $g(z) =0$ for all $z \in A$.
  4. If $f(z) g(z) =0$ for all $z \in A\cup D$, then either $f(z)=0$ for all $z \in A\cup D$ or $g(z) =0$ for all $z \in A\cup D$.

I am stuck on this problem. Can anyone help me please? where should I begin......................

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    @pankaj: I've merged the identical version of this question which you asked 5 hours ago into this one. For future reference, **don't do that**. Posting duplicate copies of the same question is rather frowned upon on this website. – 2012-12-18

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Since the zeroes of a non-zero analytic function are isolated, it may have only countably many zeroes (can you see why?).
If $f,g$ are analytic in $B\subseteq\mathbb{C}$ open, and $f(z)g(z)=0$ for all $z\in B$ then $B\subseteq N(f)\cup N(g)$ (where $N(f)$ are the zeroes of $f$). What can you say about the cardinality of $B$? could that happen if $f,g$ are both non-zero?
If $B⊆C$ is open then it is uncountable. So either $N(f)$ or $N(g)$ (or both) are uncountable. An analytic function with uncountable zeroes is zero. So either $g$ is zero or $f$ is zero, on $B$ if B is connected.
Since $A$, $D$ are connected but $A\cup D$ is not, apply the arguments to the connected components.

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    @ Dennis Gulko sir,1 may be right ,but not getting 4.. – 2012-12-19
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Hint: is $A \cup D$ connected?

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    @pankaj: Tr$y$ to grasp all that Robert is conve$y$ing in that last comment. Within, you'll find everything you need, once you unpack it. – 2012-12-20