I am confused about the property of holomorphism of complex functions.

Take the function $\frac{1}{z}$ as an example. This function satisfies the Cauchy-Riemann equations and is thus said to be holomorphic. However, doesn't the non-analytic point at $z=0$ imply that it is not holomorphic at this point? In that case, does satisfying the Cauchy-Riemann equations simply imply that the function is holomorphic in some sub-domain of the complex plane? And the property of being entire, does that then imply that it satisfies the Cauchy-Riemann equations and is differentiable on the entire complex plane?


1 Answer 1


A function is not an analytic expression. Its definition requires a domain and a codomain.When we talk about the function $\frac1z$ in the context of Complex Analysis, what we (usually) mean is the function$$\begin{array}{ccc}\mathbb C\setminus\{0\}&\longrightarrow&\mathbb C\\z&\mapsto&\dfrac1z.\end{array}$$Note that $0$ does not belong to its domain. Therefore, there is no problem with $0$. And, yes, it satisfies the Cauchy-Riemman equations at every point of its domain and it is a holomorphic function.

And this function is not entire, since its domain is not $\mathbb C$.

  • $\begingroup$ Thank you for your answer. Is there an easy way to tell if a funcion, e.g. $z+\frac{2}{z}+\frac{1}{z^3}$ is holomorphic on a domain without testing if it satisfies the Cauchy-Riemann equations? $\endgroup$
    – user766790
    Apr 4, 2020 at 15:13
  • $\begingroup$ Sure. It's the sum of holomorphic functions, right?! That's enough. $\endgroup$ Apr 4, 2020 at 15:20
  • $\begingroup$ I'm sorry, as you can see I am new to complex analysis. May I hear how you can see that $\frac{1}{z^3}$ is holomorphic? $\endgroup$
    – user766790
    Apr 4, 2020 at 15:43
  • 1
    $\begingroup$ I suppose that you agree that $z$ is holomorphic. So, $z^3$ is holomorphic, zince $z^3=z\times z\times z$. And $\frac1{z^3}$ is then the quotient of two holomorphic functions. So, it is holomorphic too. $\endgroup$ Apr 4, 2020 at 15:51
  • $\begingroup$ Ahh, right. I did not know that rule. Thank you for your help, sir. $\endgroup$
    – user766790
    Apr 4, 2020 at 15:57

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