Suppose $f$:{$z\in \mathbb C:|z|>1$}$\rightarrow\mathbb C$, $f(z)=\frac{1}{z}$. Prove that there doesn't exist an entire function $g(z)$ s.t. $f(z)=g(z)$ for all $z\in\mathbb C$ with $|z|>1$.
I really do not have any clue how to prove this. I tried it using Identity theorem but that gives me $h(z)=f(z)-g(z)$ identically zero on $\mathbb C$. Any help appreciated!! Thanks.
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You could define $\hat{f}(z)=\frac{1}{z}$ on $D=\mathbb C-\{0\}$. If $g$ existed and was equal to $f$ on $C=\{z\in\mathbb C:\|z\|>1\}$, then it'd be equal to $\hat{f}$ on $C\subset D$, so that it equals $\hat{f}$ on $D$. But there's no value you could assign to $g(0)$ that'd make this continuous at $0$.
If $g(z)=1/z$ on exterior of unit disk then it's clearly bounded there. On the other hand, $g$ must also be bounded on closed unit disk since it's continuous there. Hence $g$ is bounded entire, this is a contradiction.