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I don’t know how to proceeed. We know that the image of a Non zero entire function is dense. I was trying to use it somehow but couldn’t. Same thing happens when I tried with the Picards little theorem, that Any entire analytic function whose range omits two points must be a constant function.

Any help is appreciated..

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    $\begingroup$ Hint: Non-zero Analytic functions must have isolated zeros $\endgroup$ – Sean Nemetz Mar 6 '18 at 6:27
  • $\begingroup$ These are pretty heavy theorems. There are more basic results about differentiable functions being equal along a sequence with a limit point. $\endgroup$ – Joppy Mar 6 '18 at 6:30
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    $\begingroup$ Okay, so every point on the unit circle is a zero and they are not isolated!! Thanks a lot !!! :) @Sean Nemetz $\endgroup$ – Infinity Mar 6 '18 at 6:31
  • $\begingroup$ No problem! @infinity $\endgroup$ – Sean Nemetz Mar 6 '18 at 6:31
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Wrong theorem. The so called "identity theorem" says that if two analytic functions on a connected open set (here $f$ and the identically $0$ function) have the same values at on a set with a so called cluster or limit or accumulation point then the functions are the same (here $f=$ the identically $0$ function).

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  • $\begingroup$ The theorem you state is equivalent to $f=0$ version $\endgroup$ – user160738 Mar 6 '18 at 9:20
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Twisted proof: for $|z| < 1$, using the Cauchy integral formula: $$ f(z) = \frac1{2\pi i}\int_{|w| = 1}\frac{f(w)}{w - z}\,dw = \frac1{2\pi i}\int_{|w| = 1}\frac{0}{w - z}\,dw = 0 $$ I.e., in the unit disk $f = 0$ identically, so $\forall n\in\Bbb N$: $f^{n}(0) = 0$ and for all $z\in\Bbb C$: $$f(z) = \sum_{n=0}^\infty\frac{f^{n}(0)}{n!}\,z^n = 0.$$

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Suppose Z(f) be zero set of f.then since f is non zero entire function every pt of Z(f)is isolated pt. Hence Z(f) has to be countable. But the set |z|=1 is uncountable. Hence no such non zero entire function exists.

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