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?
 A: 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.
A: 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.
