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

Question

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..

Answer 1

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).

Answer 2

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 $z\in\Bbb C$: $$f(z) = \sum_{n=0}^\infty\frac{f^{n}(0)}{n!}\,z^n = 0.$$

Answer 3

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.