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[SOLVED] not analytic

dwsmith

Well-known member
Feb 1, 2012
1,673
Prove that there is no function $f$ such that $f$ is analytic on the punctured unit disc $\mathbb{D} - \{0\}$, and $f'$ has a simple pole at 0.

Let $f$ be analytic on the punctured disc $\mathbb{D} - \{0\}$.
$$
f(z) = \frac{g(z)}{h(z)}
$$
such that $h(z)\neq 0$.

Then
$$
f'(z) = \frac{g'(z)h(z) - g(z)h'(z)}{(h(z))^2}
$$

So $f'(z)$ has only poles of order 2. Therefore, $f'$ can't have a simple pole at 0.

How is this?
 

chisigma

Well-known member
Feb 13, 2012
1,704
Counterexample: $\displaystyle f(z)=\ln z$ is analytic in $\mathbb{D}-\{0\}$ and is $\displaystyle f'(z)=\frac{1}{z}$...

... the exact formulation of the theorem should be: prove that there is no function f(*) so that f(*) in analytic in the punctured disk $\mathbb{D}-\{0\}$, $\{0\}$ is not a brantch point, and f'(*) has a simple pole at $\{0\}$...

Kind regards

$\chi$ $\sigma$
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Counterexample: $\displaystyle f(z)=\ln z$ is analytic in $\mathbb{D}-\{0\}$ and is $\displaystyle f'(z)=\frac{1}{z}$...

... the exact formulation of the theorem should be: prove that there is no function f(*) so that f(*) in analytic in the punctured disk $\mathbb{D}-\{0\}$, $\{0\}$ is not a brantch point, and f'(*) has a simple pole at $\{0\}$...

Kind regards

$\chi$ $\sigma$
Giving that caveat. Would I be correct then?
 

chisigma

Well-known member
Feb 13, 2012
1,704

dwsmith

Well-known member
Feb 1, 2012
1,673

chisigma

Well-known member
Feb 13, 2012
1,704
If $\{0\}$ is not a brantch point, then f(*) can be written as Laurent series 'centered' in $z=0$...

$\displaystyle f(z)= \sum_{n=- \infty}^{+ \infty} a_{n}\ z^{n}$ (1)

Now if we derive (1) we obtain in any case a Laurent series in which is $a_{-1}=0$ so that f'(*) has no single pole in $\{0\}$...

Kind regards

$\chi$ $\sigma$
 

dwsmith

Well-known member
Feb 1, 2012
1,673
If $\{0\}$ is not a brantch point, then f(*) can be written as Laurent series 'centered' in $z=0$...

$\displaystyle f(z)= \sum_{n=- \infty}^{+ \infty} a_{n}\ z^{n}$ (1)

Now if we derive (1) we obtain in any case a Laurent series in which is $a_{-1}=0$ so that f'(*) has no single pole in $\{0\}$...

Kind regards

$\chi$ $\sigma$
So we can assume the series is uniformly convergent?
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Counterexample: $\displaystyle f(z)=\ln z$ is analytic in $\mathbb{D}-\{0\}$ and is $\displaystyle f'(z)=\frac{1}{z}$...

... the exact formulation of the theorem should be: prove that there is no function f(*) so that f(*) in analytic in the punctured disk $\mathbb{D}-\{0\}$, $\{0\}$ is not a brantch point, and f'(*) has a simple pole at $\{0\}$...

Kind regards

$\chi$ $\sigma$
But the theorem says f has to be analytic on the punctured disc isn't ln z not analytic on it. So the branch point part would be irrelevant.