# Complex Analysis

1. Sep 14, 2013

### nateHI

1. The problem statement, all variables and given/known data
For each of the following functions f(z), find f'(z) and identify the maximal region for which f(z) is analytic.

1. $f(z)=1/(z^2+1)$
2. $f(z)=e^{-1/z}$

2. Relevant equations

3. The attempt at a solution
1. $f'(z)=\frac{-2z}{(z^2+1)^2}$ <--this part is easy. I'm having difficulty being certain of the maximum region for analyticity. Here is my attempt.

f(z) is analytic everywhere but + or - i because f'(z) is undefined there.

Is that a true stament or is the correct statement ... f(z) is analytic everywhere but + or - i because f(z) is undefined there.

2. $f'(z)=\frac{e^{-1/z}}{z^2}$ <--this part is easy. I'm having difficulty being certain of the maximum region for analyticity. Here is my attempt.

f(z) is analytic everywhere but 0 because f'(z) is undefined there. However, f(z) is analytic at infinity.

Is that a true stament or is the correct statement ... f(z) is analytic everywhere but 0 because f(z) is undefined there. However, f(z) is analytic at infinity.

2. Sep 15, 2013

### Tsunoyukami

The short answer is that you are correct.

In the future the simplest way to approach these problems is to remember the definition of analytic:

Definition: A function $f(z)$ is analytic at a point $z_{o}$ if $lim_{z \rightarrow z_{o}} \frac{f(z) - f(z_{o})}{z - z_{o}} = lim_{h \rightarrow 0} \frac{f(z_{o} + h) - f(z_{o})}{h}$.

The maximal region for which $f(z)$ is analytic will be the entire complex plane with any singularities removed (read: with the places it is undefined removed.

For example, for your second function we can write:

The function $f(z) = e^{\frac{-1}{z}}$ is analytic on ℂ - {0}.

3. Sep 15, 2013

### vela

Staff Emeritus
The first statement is the one you want. There's a theorem that says if a function is complex differentiable at a point, it's analytic at that point. Just because a function exists at a point doesn't mean it's analytic there.

4. Sep 16, 2013

### jackmell

I don't think so huh vela? Doesn't it have to be complex-differentiable in some disc centered at the point in order for it to be analytic at that point?

5. Sep 16, 2013

### vela

Staff Emeritus
Yeah, you're right.