Question about branch of logarithms

[MathLearner123]

I've read a proof from Complex Made Simple (David C. Ullrich)
Proposition 4.3. Suppose that $V$ is an open subset of the plane. There exists a branch of the logarithm in $V$ if and only if there exists $f \in H(V)$ with $f'(z) = \frac{1}{z}$ for all $z \in V$.

Proof: One direction is trivial and the other is very easy: Suppose that $f \in H(V)$ and $f'(z) = \frac{1}{z}$ for all $z \in V$. Choose $z_0 \in V$. Note that $z_0 \ne 0$, so $z_0$ has a logarithm. Hence we can choose a constant $c \in \mathbb{C}$ such that $e^{c+f(z_0)} = z_0$, and now $L$ is a branch of the logarithm in $V$, if $L(z) = c+f(z)$ for all $z \in V$.

I don't understand how we can choose that constant $c \in \mathbb{C}$ and how it is a branch of logarithm if we know that $e^{L(z)} = z$ only for $z = z_0$.

Thanks!

 

[FactChecker]

Suppose two functions, $f, g \in H(V)$ agree at a single point and have identical derivatives in $V$. What can you conclude?

 

[MathLearner123]

Suppose two functions, $f, g \in H(V)$ agree at a single point and have identical derivatives in $V$. What can you conclude?

I don't know if they are equal if $V$ is not connected.

 

[FactChecker]

I don't know if they are equal if $V$ is not connected.

Very good point. I stand corrected. In addition to that, I'm not sure that my comment gets you any farther. It just seems that the derivative $1/z$ has not been adequately used yet.

 

[MathLearner123]

Very good point. I stand corrected. In addition to that, I'm not sure that my comment gets you any farther. It just seems that the derivative $1/z$ has not been adequately used yet.

I think that the statement is incorrect, and $V$ needs to be a domain (i.e. connected and open set), and this will make the statement true. But I still waiting for answers. I really need an answer

 

[WWGD]

IIRC, we need a connected , simply-connected set, so that $\int_{\gamma}\frac{dz}{z}$
Is well-defined, i.e., independent of path.

 

[WWGD]

I've read a proof from Complex Made Simple (David C. Ullrich)
Proposition 4.3. Suppose that $V$ is an open subset of the plane. There exists a branch of the logarithm in $V$ if and only if there exists $f \in H(V)$ with $f'(z) = \frac{1}{z}$ for all $z \in V$.

Proof: One direction is trivial and the other is very easy: Suppose that $f \in H(V)$ and $f'(z) = \frac{1}{z}$ for all $z \in V$. Choose $z_0 \in V$. Note that $z_0 \ne 0$, so $z_0$ has a logarithm. Hence we can choose a constant $c \in \mathbb{C}$ such that $e^{c+f(z_0)} = z_0$, and now $L$ is a branch of the logarithm in $V$, if $L(z) = c+f(z)$ for all $z \in V$.

I don't understand how we can choose that constant $c \in \mathbb{C}$ and how it is a branch of logarithm if we know that $e^{L(z)} = z$ only for $z = z_0$.

Thanks!

The issue with $z_0$ is that it's arbitrary, and as long as $z_0 \neq 0$, the rest will hold. I assume we define a log of $z$ to be a holomorphic function $f(z)$ with $e^{f(z)}=z$. Then $\frac{d}{dz}(e^{f(z)})=e^{f(z)}f'(z)=1$, so that $f'(z)=\frac{1}{z}$
Edit: The remainder of the argument involves the use of an auxiliary function $g(z):=ze^{-f(z)}$, differentiate it, show it's constant, and the rest will follow.

 

[MathLearner123]

@WWGD if derivative of $g$ is constant, I still need to have that $V$ is domain to show that. So I also need that $V$ is connected.

 

[FactChecker]

@WWGD if derivative of $g$ is constant, I still need to have that $V$ is domain to show that. So I also need that $V$ is connected.

If $V$ is not connected, you can still define a branch of a logarithm in each connected, open part of $V$. The derivative is still 1/z in each part. The definitions will not be unique, but the original statement does not insist on a unique solution.

 

[MathLearner123]

If $V$ is not connected, you can still define a branch of a logarithm in each connected, open part of $V$. The derivative is still 1/z in each part. The definitions will not be unique, but the original statement does not insist on a unique solution.

Oh so it will be a branch of logarithm in each connected, open part, and it will be something like a piecewise branch of logarithm on $V$?

 

[mathwonk]

And note that, as you realize, the proof as given is incorrect, as it assumes V is connected; so the argument, and the choice of a fixed point z0, and constant c, should be extended to separate such choices in every connected component of V. I.e., as stated, the function L(z) will usually not be a logarithm in any connected component other than the one containing z0.

I am just stating explicitly what you and Factchecker have pointed out.

 

[WWGD]

Not just connected, but simply-connected, for the logarithmic integral to be well-defined. The standard motivation is $\int_ {\gamma} \frac{dz}{z}=i2\pi \neq 0$

 

[FactChecker]

Oh so it will be a branch of logarithm in each connected, open part, and it will be something like a piecewise branch of logarithm on $V$?

I am not sure if all the normal properties of the logarithm can remain true unless the definitions in different sections are compatible.

 

[FactChecker]

Not just connected, but simply-connected, for the logarithmic integral to be well-defined. The standard motivation is $\int_ {\gamma} \frac{dz}{z}=i2\pi \neq 0$

The OP specified that there would be a branch of the logarithm. I assumed that it would allow a branch cut and the associated discontinuity at the branch cut for any part of the domain that encircled z=0.

 

[WWGD]

The OP specified that there would be a branch of the logarithm. I assumed that it would allow a branch cut and the associated discontinuity at the branch cut for any part of the domain that encircled z=0.

I think the condition generalized to simply-connected regions, as the issue is that of winding around points not in the region.