Question regarding holomorphic functions

[Susanne217]

Homework Statement

Given a complex valued function f(z) = 1/z^2+1 show the area for which its holomorphic?

Homework Equations

I know that if f:\Omega \rightarrow \mathbb{C} and z_0 \in \Omega

then f'(z_0) = \lim_{z \rightarrow z_0} \frac{f(z)-f(z_0)}{z-z_0}

if the limit exists then f is holomorphic at the point z_0...

The Attempt at a Solution

To show the area for which f is holomorphic isn't this simply to check if the definition above can be applied to every z_0 of f?? Or am I missing something here?

where the two possiblites for z_0 = \pm i or is it simply that f is holomorphic on the area \Omega - \{\pm i\} ??

Best Regards
Susanne

 

[HallsofIvy]

First, is that (1/z^2)+ 1= \frac{1}{z^2}+ 1 or 1/(z^2+ 1)= \frac{1}{z^2+ 1}[/itex]?<br /> <br /> The first is not defined at z= 0 and the second is not defined at z= i or z= -i.

 

[Susanne217]

First, is that (1/z^2)+ 1= \frac{1}{z^2}+ 1 or 1/(z^2+ 1)= \frac{1}{z^2+ 1}[/itex]?<br /> <br /> The first is not defined at z= 0 and the second is not defined at z= i or z= -i.

<br /> <br /> HallsofIvy its suppose to to be <br /> <br /> f(z) = \frac{1}{z^2+1}<br /> <br /> Isn&#039;t the point being here that the holomorphic definition can be applied to every point in \Omega except \pm i ??

 

[Susanne217]

HallsofIvy its suppose to to be

f(z) = \frac{1}{z^2+1}

Isn't the point being here that the holomorphic definition can be applied to every point in \Omega except \pm i ??

By that I mean that according to the definition of holomorph then a function can only be called holomorphic iff its complex differentiable in all points...

But since f&#039;(z_0 = \pm i) doesn't exist then as I understand the definition of Holomorphic functions that f is holomorphic \forall z_0 \in \mathbb{C} \setminus \{\pm i\}

Because (\frac{f}{g}^{\prime})(z_0) = 0 then z_0 = \pm i and hence the definition of Holomorphic doesn't apply then z_0 = \pm i. Then the area for which f is holomorphic is \forall z_0 \in \mathbb{C} \setminus \{\pm i\}.. As I see it...

where \Omega(z_0,r) where r>0 with the exception above is the area for which f is holomorphic...

How is that Hallsoftivy??