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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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For some basic information about writing math at this site see e.g. here, here, here and here. I tried to improve your post using TeX (for better readability). Please check whether these edits did not unintentionally change the meaning of your post. – Julian Kuelshammer Dec 18 '12 at 9:06
How many zeroes can an analytic non-zero function have? – Dennis Gulko Dec 18 '12 at 13:36
But could it be uncountable? – Dennis Gulko Dec 18 '12 at 13:41
@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. – Willie Wong Dec 18 '12 at 13:52

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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I edited my answer. Please try to remove unnecessary comments, to shorten the thread. I'm trying to lead you all the way to the solution, but you must understand what's going on by yourself. It will be totally worthless to give you the complete solution. – Dennis Gulko Dec 19 '12 at 14:35
@pankaj: Obsereve that if $f(z)g(z)=0$ for all $z\in A\cup D$ then, in particular $f(z)g(z)=0$ for all $z\in A$... – Dennis Gulko Dec 19 '12 at 14:37
@ Dennis Gulko sir, we can't apply the arguments to A∪D, so, what about 1 and 4??? – pankaj Dec 19 '12 at 14:45
@ Dennis Gulko sir,1 may be right ,but not getting 4.. – pankaj Dec 19 '12 at 15:00

Hint: is $A \cup D$ connected?

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If they are connected, there must be some $z$ that is in their intersection. What $z$ is it? – tacos_tacos_tacos Dec 18 '12 at 8:56
@pankaj: I think Robert may be waiting for you to figure out the correct answer to the question he posed you. – Willie Wong Dec 18 '12 at 13:58
z that is in their intersection is z=1 – pankaj Dec 18 '12 at 13:59
No, $1$ is in neither $A$ nor $D$. – Robert Israel Dec 18 '12 at 19:56
$A$ being a connected open set, if $fg = 0$ there one of the two must be $0$ on all of $A$. Same for $D$. But $A \cup D$ is not connected, so what happens on $A$ does not influence what happens on $D$. You could have, say, $f = 0$ on $A$ but not $D$, $g = 0$ on $D$ but not $A$. – Robert Israel Dec 19 '12 at 19:06

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