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let $f$ is analytic function on domain $D$.if $B(a,R)$ is open ball inside $D,$ such that $f=0,\forall z\in B(a,R)$. then does it imply $f $ is zero on full $D$. please explain

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    This follows from the identity theorem in complex analysis, if I'm not mistaken. – 2017-02-14
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    [Yes, this is true.](https://en.wikipedia.org/wiki/Principle_of_permanence) – 2017-02-14

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I reproduce it here as I like it very much. Consider the null set of $f$, it is closed (as $f$ is continuous) with non-empty interior as your ball is inside it and call $N$ the connected component which contains $a$ of this interior. If $N=D$, we are done, otherwise, consider any point of its boundary, develop $f$ in Taylor series around this point and see that this neighbourhood is in $N$. Then $N$ is closed and open. As $N\not= \emptyset$, we must have $N=D$ and $f\equiv 0$.

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    Thank you. Hope all is cristal clear and helps. – 2017-02-14
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That is a consequence of the identity theorem for analytic functions. For a proof see:

https://en.wikipedia.org/wiki/Identity_theorem