# Is $f(z)=\frac{1}{z}$ an entire function?

Of course $f(z)=\frac{1}{z}$ is not an entire function since following limit doesn't exists. $$\lim_{z \to 0} \frac{\frac{1}{z} - 0}{z-0} =\lim_{z \to 0} \frac{1}{z^2}=\infty$$

However, if I take $\mathbb{C} \cup\{\infty\}$, as a domain and range of given function, and define like this, $$\frac1{\infty}=0$$ $$\frac1{0}=\infty$$ then is $\dfrac{1}{z}$ a entire function?

• If you have that domain, which functions aren't entire?
– Mark
Aug 14 '16 at 10:30
• @Mark The ones with essential singularities, like $e^z$. Aug 14 '16 at 10:32
• I think i choose $\mathbb{C}$ as a domain and range, then it's not a entire function and letting $\bar{\mathbb{C}}$ as a domain and range will make 1/z entire function Aug 14 '16 at 10:34
• If you're going to change the domain, you have to come up with a definition of "entire function" before you can ask whether something is one. Aug 14 '16 at 10:52
• The usual notion of entire function on $\mathbb{C} \cup \{ \infty \}$ is such that the only entire functions are the constant ones.
– quid
Aug 14 '16 at 11:06

In complex analysis (and complex geometry), an "entire function" usually refers to a complex-valued function. Since the reciprocal mapping $f(z) = 1/z$ (extended to the Riemann sphere/complex projective line) is not complex-valued ($f(0) = \infty$ is not a complex number), one doesn't usually call $f$ "entire".
This mapping $f$ is, however, a (global) holomorphic mapping of the sphere to itself, in fact, a projective automorphism (a holomorphic bijection with holomorphic inverse). In this sense, your intuition is perfectly correct.