Where is ##(z+1)Ln(z)## differentiable?

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Homework Help Overview

The discussion revolves around determining the domain in which the complex function ##f(z)=(z+1)Ln(z)## is differentiable, with a focus on the properties of the principal complex logarithmic function ##Ln(z)##.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to express ##z## in terms of its real and imaginary components and apply the Cauchy-Riemann equations. Questions arise regarding the simplification of ##Arg(z)## within the context of the logarithmic function.

Discussion Status

Participants are exploring different interpretations of the logarithmic function and its components. Some guidance has been offered regarding the expression for ##Arg(z)##, but there is still uncertainty about reconciling these expressions with established resources.

Contextual Notes

There is mention of potential typos and confusion regarding the definitions and conditions associated with the argument of a complex number, as well as the continuity of the real and imaginary parts of the function.

Terrell
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Homework Statement


Find the domain in which the complex-variable function ##f(z)=(z+1)Ln(z)## is differentiable. Note: ##Ln(z)## is the principal complex logarithmic function.

Homework Equations


Cuachy-Riemann Equations?

The Attempt at a Solution


The solution I have in mind would be to let ##z=x+iy## then substitute and simplify. Check if it satisfies the Cauchy-Riemann equations, the real and imaginary part of ##f## is continuous and their first-order partial derivative are continuous as well. But, I do not know how to simplify ##Arg(z)## in ##Ln(z)=Log_e(z)+iArg(z)## because ##z## is not fixed.
 
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It should be ##\log |z|## and not ##\log z##, i.e. natural logarithm of the modulus of z.
 
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MathematicalPhysicist said:
It should be ##\log |z|## and not ##\log z##, i.e. natural logarithm of the modulus of z.
Yes, I made a typo, but how do I simplify ##Arg(z)##?
 
Well, ##Arg(z)=\arctan y/x## where ##z=x+iy##, this should help you with Cauchy-Riemann.
 
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MathematicalPhysicist said:
Well, ##Arg(z)=\arctan y/x## where ##z=x+iy##, this should help you with Cauchy-Riemann.
It's what I have on paper but I can't reconcile with what wikipedia have. It has conditions depending on the values of x and y. https://en.wikipedia.org/wiki/Argument_(complex_analysis)
 
Terrell said:
It's what I have on paper but I can't reconcile with what wikipedia have. It has conditions depending on the values of x and y.
After giving it some thought now, it doesn't seem to matter when I start differentiating.
 

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