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 [tex]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.[/tex]

 

[Susanne217]

First, is that $$(1/z^2)+ 1= \frac{1}{z^2}+ 1$$ or [tex]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.[/tex]

[tex] <br /> HallsofIvy its suppose to to be <br /> <br /> $$f(z) = \frac{1}{z^2+1}$$<br /> <br /> Isn't the point being here that the holomorphic definition can be applied to every point in $$\Omega$$ except $$\pm i$$ ??[/tex]

 

[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 $$f'(z_0 = \pm i)$$ 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??