Complex analysis - the logarithmic function

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Homework Help Overview

The problem involves demonstrating that the function Log(-z) + i(pi) is a branch of log(z) that is analytic in the domain D*, which excludes points on the nonnegative real axis. The context is within complex analysis, specifically focusing on the properties of logarithmic functions.

Discussion Character

  • Exploratory, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the analytic nature of Log(-z) and its implications for Log(-z) + i(pi). There is an exploration of the relationship between the exponential function and the logarithmic function, particularly in terms of branch cuts and analytic continuation.

Discussion Status

Some participants have provided insights into the analytic properties of the logarithmic function and its branches. There is an ongoing exploration of how to formally demonstrate that Log(-z) + i(pi) qualifies as a branch of log(z) through the use of the exponential function.

Contextual Notes

Participants are navigating the definitions and properties of logarithmic functions in complex analysis, particularly focusing on branch cuts and the conditions under which these functions remain analytic.

mariab89
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Homework Statement



Show that the function Log(-z) + i(pi) is a branch of logz analytic in the domain D* consisting of all points in the plane except those on the nonnegative real axis.

Homework Equations





The Attempt at a Solution



I know that log z: = Log |z| + iArgz + i2k(pi)
I'm not sure where to start with this question, any help would be greatly appreciated!
thanks :)
 
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You know Log(z) is analytic except for a branch cut on the negative real axis. That means Log(-z) is analytic except for a branch cut on the positive real axis. So is Log(-z)+i*pi. Now you just have to show that it is a branch of log(z) by showing exp(Log(-z)+i*pi)=z.
 
so to show that what i did was...

exp(Log(-z) + i*pi) = exp(Log(-z))exp(i*pi) = (-z) (-1) = z

but.. I am still unclear how this shows that Log(-z) + i*pi is a branch of log z.
 
If g(z) satisfies exp(g(z))=z then it's a branch of log(z). That's what defines log(z). It's an inverse function of exp(z).
 
oh ok i see now!
Thanks a lot!
 

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