How many entire functions are equal to $\frac{1}{z}$ for $|z| > 1$? Let $f:\{z\in\mathbb{C}:|z|>1\}\rightarrow\mathbb{C},f(z)=\frac{1}{z}.$ Now my question is how many entire functions $g$ are there such that $f=g$ for $|z|>1$?According to me there is no such entire function because if there is so then it must be equal to $\frac{1}{z}$ itself. Am I right? Please help. Thanks.
 A: Let $C$ be the circle centered at $0$ and of radius $2$ (which is included in the domain of $f$). Now, compute:
$$\oint_C f(z)\,\mathrm{d}z=2i\pi\neq0.$$
Hence, by Goursat's Theorem (aka Cauchy's Integral Theorem), there are no entire functions $g$ such that $f$ and $g$ coincide on $C$.
A: Your argument is incomplete, but can be improved by using the identity theorem to deduce that for any such entire function $g$, it must satisfy $g(z) = \frac{1}{z}$ for $z\neq 0$. However, it still needs to be shown that no such $g$ exists (maybe you've done this already, in which case, you have the result). One way to prove this is to compute an integral similar to the one below. However, if you are going to compute such an integral, you can actually arrive at the result directly without having to apply the identity theorem.

Suppose there was entire function $g$ such that $f = g$ on $\{z \in \mathbb{C} \mid |z| > 1\}$. Consider the path given by $\gamma : [0, 2\pi] \to \mathbb{C}$, $\gamma(t) = 2e^{it}$. The image of $\gamma$ is the circle centred at the origin of radius $2$. Then
$$\int_{\gamma}g(z)dz = \int_{\gamma}f(z)dz = \int_0^{2\pi}f(\gamma(t))\gamma'(t)dt = \int_0^{2\pi}\frac{1}{2e^{it}}2ie^{it}dt = \int_0^{2\pi}idt = 2\pi i,$$
but as $g$ is holomorphic on the disc of radius $2$ centred at the origin, $\int_{\gamma}g(z)dz = 0$, which is a contradiction.
A: Yes, you're correct. The function must by the identity theorem coincide on the domain of $f$. And there's no entire function such that $g(z)=1/z$ for $z\ne 0$. To see this you use the fact that $g$ has to be continuous ond therefore $g(0)=\infty$, but it has to be derivable there which would not work.
