Does there exists a non-zero entire function f such that $f(z)=0$ , $|z|=1$?

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..

• Hint: Non-zero Analytic functions must have isolated zeros – Sean Nemetz Mar 6 '18 at 6:27
• These are pretty heavy theorems. There are more basic results about differentiable functions being equal along a sequence with a limit point. – Joppy Mar 6 '18 at 6:30
• Okay, so every point on the unit circle is a zero and they are not isolated!! Thanks a lot !!! :) @Sean Nemetz – Infinity Mar 6 '18 at 6:31
• No problem! @infinity – Sean Nemetz Mar 6 '18 at 6:31

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).
• The theorem you state is equivalent to $f=0$ version – user160738 Mar 6 '18 at 9:20
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 $z\in\Bbb C$: $$f(z) = \sum_{n=0}^\infty\frac{f^{n}(0)}{n!}\,z^n = 0.$$