Finding Holomorphic Logarithmic Formulas for Half-Planes in Complex Analysis

Click For Summary

Homework Help Overview

The discussion revolves around finding holomorphic logarithmic formulas for complex numbers in two distinct half-planes: one above the x-axis and one below. Participants are tasked with demonstrating the holomorphic nature of the logarithm in these regions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the definition of the logarithm in the context of the upper and lower half-planes, questioning how to ensure it is well-defined and holomorphic. There is mention of using the Cauchy-Riemann equations and the definition of the derivative to explore these properties.

Discussion Status

The conversation is active, with participants exploring various aspects of the problem, including the implications of defining the logarithm in closed half-planes and the continuity of angle functions. Some guidance has been offered regarding the use of the Cauchy-Riemann equations, but no consensus has been reached on the specifics of the approach.

Contextual Notes

Participants express uncertainty about whether the problem refers to open or closed half-planes, particularly concerning the definition of the logarithm at zero. This ambiguity influences their reasoning and approach to the problem.

Metric_Space
Messages
96
Reaction score
0

Homework Statement




i) Find a suitable formula for log z when z lies in the half-plane K that lies above the x-axis, and
from that show log is holomorphic on K

ii) Find a suitable formula for log z when z lies in the half-plane L that lies below the x-axis, and
from that show log is holomorphic on L.

Homework Equations





The Attempt at a Solution



I've found a formula log Z on wikipedia but not sure how to relate it to the half plane(s).
 
Physics news on Phys.org
Ok, so what's that formula for log(z)? You only have to worry about defining a continuous angle function in the half planes, right?
 
I think it is:

Log z: = ln r + iθ = ln | z | + iArg z.
 
Metric_Space said:
I think it is:

Log z: = ln r + iθ = ln | z | + iArg z.

So, this is well-defined in the upper-half plane right? Can you show it's also holomorphic there?
 
I think it's well-defined on the upper-half plane.

Not sure how to show it's holomorphic -- should I just substitute into the definition of derivative?
 
micromass said:
So, this is well-defined in the upper-half plane right? Can you show it's also holomorphic there?

I wonder if they mean the closed half-plane? The problem would be a little more challenging if they do.
 
Dick said:
I wonder if they mean the closed half-plane? The problem would be a little more challenging if they do.

Good point. But is it even true in the closed half plane? You can easily pick your line of discontinuity to lie under the x-axis, but you cannot define the logarithm of 0 in any satisfying way, can you?
 
Metric_Space said:
I think it's well-defined on the upper-half plane.

Not sure how to show it's holomorphic -- should I just substitute into the definition of derivative?

You could do that. But if I were you, I would try the Cauchy-Riemann equations though. Maybe you can even use the inverse function theorem...
 
micromass said:
Good point. But is it even true in the closed half plane? You can easily pick your line of discontinuity to lie under the x-axis, but you cannot define the logarithm of 0 in any satisfying way, can you?

Good point also. Guess I was thinking closed half plane\{0}. So never mind.
 
  • #10
how would I use the C-R equations in this case?
 
  • #11
Just find the partial derivatives, show they are continuous and show that the CR-equations are satisfied. This would imply holomorphicness...
 
  • #12
great
 
  • #13
you have to change co-ordinates from /9x,y) to (r,theta), this is done via the chain rule.
 

Similar threads

Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
8
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 14 ·
Replies
14
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K