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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?

Lorenzo B.
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MrTanorus
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  • If you have that domain, which functions aren't entire? – Mark Schultz-Wu Aug 14 '16 at 10:30
  • @Mark The ones with essential singularities, like $e^z$. – Arthur 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 – MrTanorus Aug 14 '16 at 10:34
  • @arthur I know, it just removes a whole class of meromorphic functions, which I was trying to point out. With regards to this problem, if you work on the Riemannian sphere (complex plane plus infinity), you need to take into account singularities at infinity as well. Notably, things like polynomials now have singularities, which may not be obvious – Mark Schultz-Wu Aug 14 '16 at 10:35
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    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. – Gerry Myerson Aug 14 '16 at 10:52
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    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

2 Answers2

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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.

In the same sense, every rational function in one variable defines a global holomorphic mapping of the sphere to itself, and the rational functions of degree one (i.e., Möbius transformations) are precisely the holomorphic automorphisms of the sphere.

Andrew D. Hwang
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Since the Entire function is defined to be Analytic everywhere in the complex plane and we see the function f(z)=1/z is not Analytic everywhere in the complex plane because the f(z) is not analytic at z=0 thats why f(z) is not en Entire function.

Rana Ryhan
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