Question about branch of logarithms

  • #1
MathLearner123
5
2
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!
 
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  • #2
Suppose two functions, ##f, g \in H(V)## agree at a single point and have identical derivatives in ##V##. What can you conclude?
 
  • #3
FactChecker said:
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.
 
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  • #4
MathLearner123 said:
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.
 
  • #5
FactChecker said:
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 :smile:
 
  • #6
IIRC, we need a connected , simply-connected set, so that ##\int_{\gamma}\frac{dz}{z}##
Is well-defined, i.e., independent of path.
 
  • #7
MathLearner123 said:
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.
 
Last edited:
  • #8
@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.
 
  • #9
MathLearner123 said:
@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.
 
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  • #10
FactChecker said:
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##?
 
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  • #11
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.
 
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  • #12
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##
 
  • #13
MathLearner123 said:
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.
 
  • #14
WWGD said:
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.
 
  • #15
FactChecker said:
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.
 

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