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Branches of the complex logarithm

  • Thread starter JonathanT
  • Start date
18
0
1. Homework Statement
Find a branch of log(z − 1) that is analytic inside the unit circle. What is the value of this branch at z = 0?

2. The attempt at a solution

Alright so clearly the log(z) function is analytic at all points accept for the negative real axis.

So log(z-1) will be analytic at all points x ≤ 1. My problem is choosing a branch. I'm really bad at doing this. I understand if the function was something like log(z+1) it would already be analytic inside the unit circle using the principle branch. I just don't see how I can choose a branch that is analytic inside the unit circle. I think if I understand this problem I can be prepared for my test tomorrow. Thanks for any help in advance.
 
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I just don't see how I can choose a branch that is analytic inside the unit circle.
As just the real axis is problematic, you can begin in the upper [or lower] part of the plane, choose a branch there, and try to extend this to the interior of the unit circle afterwards.
 
18
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So if I understand right from what I've got in my notes I can just start my branch at some arbitrary point say something like.

[itex]\tau[/itex]<arg(w)≤[itex]\tau[/itex]+2[itex]\pi[/itex]

Where [itex]\tau[/itex] = [itex]\pi[/itex]/4 maybe?

However, then the function would not be analytic at the angle [itex]\tau[/itex] on the unit circle am I right?
 
18
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Tell me if this works:

log(z-1) = log(-1) + log(1-z) = Log|-1| + i[itex]\cdot[/itex]arg(-1) + log(1-z)

= 0 + i[itex]\cdot[/itex][itex]\pi[/itex] + log(1-z)

So I end up with f(z) = log(1-z) + i[itex]\cdot[/itex][itex]\pi[/itex]

Then I could use the principle argument because now the cut is at real values of x ≥ 1. Which is outside the inside of the unit circle. I have no idea if my logic here is right though.
 
33,382
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That should work, right.
 
18
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Awesome! I think I finally get this. Hope I do well on the test today. Thanks!
 

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