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"If we should speak of a function $f$ that is analytic in a set $S$ which is not open, it is to be understood that $f$ is analytic in an open set $S'$ containing $S$." Is it correct to illustrate this in a broken (dotted/dashed) circle with center $z_0$ (for example) that is inside another bigger circle with boundary not dashed/broken?

Also, "f is analytic at a point $z_0$ if it is analytic in a neighborhood of $z_0$". In the function $f(z)= 1/z$, Say $z_0 = 0$, the derivative of f exists elsewhere other than zero, or its neighboring points. But $f(z)$ is not analytic at $0$. Is the statement not applicable on this or I got it wrong? I am confused. Thanks.

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By definition, a neighbourhood of $0$ contains $0$. Your example $1/z$ is not analytic in a neighbourhood of $0$. The set where it is analytic is called a deleted neighbourhood of $0$.

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  • $\begingroup$ How about the first question, sir? Can I represent the statement that way? $\endgroup$ – frstrtdmthmtcn Nov 3 '17 at 3:36
  • $\begingroup$ Sir, what is the implication if a certain function's derivative exists only at a point z=0? $\endgroup$ – frstrtdmthmtcn Nov 3 '17 at 5:16

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