Let $f(z) = 1/z$ be defined on $\{ z \in \Bbb{C} : |z| > 1\}$. Then can we find an entire function $g$ such that $f(z) = g(z)$ for $|z| > 1$?

Question

Let $f(z) = 1/z$ be defined on $\{ z \in \Bbb{C} : |z| > 1\}$. Then can we find an entire function $g$ such that $f(z) = g(z)$ for $|z| > 1$?

Now, can we plainly apply uniqueness theorem and say that such a function $g$ can not exist?

Answer 1

If such a function $g$ existed, its integral over any closed curve would be zero by Cauchy's theorem. However,

$\oint_{|z|=2}g(z)dz$

is obviously nonzero.

Answer 2

Suppose $\;g(z)\;$ is entire (i.e. analytic in the whole complex plane) and $\;g(z)=\dfrac1z\;$ on $\;|z|>1\;$ .

Since $\;\dfrac1z\;$ is analytic on $\;0<|z|\le1\;$ as well, $\;g(z)$ continues analitically $\;\dfrac1z\;$ on the punctured unit disk.

But since for any $\;z_k\to 0\;$ have that $\;g(z_k)\to g(0)\;$ and $\;g(0)\;$ is well defined, we get that $\;\dfrac1z\;$ can be analitically continued on $\;z=0\;$ as well , which of course is absurd.