Let P=C/{1} and Q=mod(z)>2 and f(z) is analytic on P, then there does not exists a non constant analytic function P to Q .

Question

Here domain is C/{0} and codomain is mod(z)>2 i have to claim that f(z) is constant , now my intuition is if i prove that f(z) has a Removable singularity at z=1(if exists) then it can be extended on C analytically so , f(z) can be redefine at the point z=1 , then by picard's theorem a function f(z) can leave atmost one value of Complex sets from it's range but here codomain is mod(z)>2 , so function has to be constant (plz tell if my intuition is correct) , now iam having difficulties to show why f(z) if singularity exists at z=1 is a Removable singularity , i know since f(z) is analytic on C/{1} so singularity if exists is an isolated singularity and isolated singularity is of 3 types - (1) removable , (2) pole and (3) is essential , if it has essential singularity at z=1 then in the deleted neighbourhood of z=1 f(z) can be made closer to any complex value ( and leave atmost 1) but codomain is mod(z)>2 so uncountable many points inside the disk mod(z)<=2 will not be in it's range so essential singularity is not possible, how can we say f(z) does not have have a pole but a Removable singularity ?? i know*from theory) if z=1 is a Removable singularity of f(z) at z=1 then f(z) is bounded in some neighbourhood of z=1 , but i can't able to see this.

Answer 1

Just consider the function $ g(z) = \frac {1}{f(z)}$ . See that since |f(z)| > 2 , g is bounded around a neighborhood of 1 and hence it has a removable singularity at 1 . So g becomes an entire function with image contained inside the disk of radius 2 . So by $ \textit{Picard's Theorem }$ g must be constant . So f is constant .