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If $f:\Bbb C \to \Bbb C$ is continuous such that $f$ is analytic outside the bounded set $S$ then prove that $f$ is entire.

To prove this we have to prove that $f$ is analytic in $S$. For this I I think it can be done by using Morera's theorem , which states that " If $f$ is continuous in a simply connected domain $D$ and $\int_Cf(z)\,dz=0$ , for every closed contour $C$ in $D$ then $f$ is analytic in $D$."

But here how I can show that $\int_C f(z)\,dz=0$ for every closed contour $C$ in the bounded domain $S$?

I've seen this similar type question , but there are no enough proof which I want. So , please don't make this question as a duplicate question , and give me some hints to show the integral value is $0$.

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    $\begingroup$ What are the conditions on $S$? Is it a line segment? This is definitely not true for arbitrary bounded set $S$. $\endgroup$ – Owen Sizemore Mar 31 '17 at 20:09
  • $\begingroup$ @OwenSizemore Actually it is given just a Bounded Set. I don't know whether it is correct or not. If you think there re something wrong then clarify it , please. $\endgroup$ – Empty Mar 31 '17 at 20:11
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This is not true. To see this consider the following counterexample.

$$f(x) = \begin{cases} z \textrm{ if } |z| \geq 1 \\ z\cdot e^{i\cdot(|z|-1)} \textrm{ if } \frac{1}{2} \leq |z| < 1 \\ z\cdot e^{i\cdot|z|} \textrm{ if } |z| < \frac{1}{2} \end{cases} $$

It's not hard to see that this function is continuous everywhere and analytic outside of the unit circle but is not analytic inside the continuous circle.

Basically the idea is that if the set $S$ contains an open set then you have enormous freedom to make the function because being continuous inside this set is a fairly weak condition while being analytic is a $\textit{very}$ strong condition.

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