Complex Analysis: Find f'(z) & Region of Analyticity

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Homework Help Overview

The discussion revolves around finding the derivative of complex functions and identifying their regions of analyticity. The functions in question are f(z) = 1/(z^2 + 1) and f(z) = e^{-1/z}, both of which involve considerations of singularities and points where the functions may not be analytic.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the derivatives of the functions and express uncertainty regarding the maximal regions of analyticity. Questions arise about the correctness of statements regarding where the functions are analytic and the implications of complex differentiability.

Discussion Status

The discussion includes attempts to clarify the definitions and implications of analyticity, with some participants providing guidance on the relationship between complex differentiability and analyticity. Multiple interpretations of the conditions for analyticity are being explored.

Contextual Notes

Participants are navigating the definitions and theorems related to complex functions, particularly focusing on singularities and the conditions under which functions are considered analytic. There is an emphasis on understanding the nature of singular points and the implications for the functions' behavior in the complex plane.

nateHI
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Homework Statement


For each of the following functions f(z), find f'(z) and identify the maximal region for which f(z) is analytic.

1. f(z)=1/(z^2+1)
2. f(z)=e^{-1/z}

Homework Equations





The Attempt at a Solution


1. f&#039;(z)=\frac{-2z}{(z^2+1)^2} <--this part is easy. I'm having difficulty being certain of the maximum region for analyticity. Here is my attempt.

f(z) is analytic everywhere but + or - i because f'(z) is undefined there.

Is that a true stament or is the correct statement ... f(z) is analytic everywhere but + or - i because f(z) is undefined there.

2. f&#039;(z)=\frac{e^{-1/z}}{z^2} <--this part is easy. I'm having difficulty being certain of the maximum region for analyticity. Here is my attempt.

f(z) is analytic everywhere but 0 because f'(z) is undefined there. However, f(z) is analytic at infinity.

Is that a true stament or is the correct statement ... f(z) is analytic everywhere but 0 because f(z) is undefined there. However, f(z) is analytic at infinity.
 
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The short answer is that you are correct.

In the future the simplest way to approach these problems is to remember the definition of analytic:

Definition: A function ##f(z)## is analytic at a point ##z_{o}## if ##lim_{z \rightarrow z_{o}} \frac{f(z) - f(z_{o})}{z - z_{o}} = lim_{h \rightarrow 0} \frac{f(z_{o} + h) - f(z_{o})}{h}##.The maximal region for which ##f(z)## is analytic will be the entire complex plane with any singularities removed (read: with the places it is undefined removed.

For example, for your second function we can write:

The function ##f(z) = e^{\frac{-1}{z}}## is analytic on ℂ - {0}.
 
nateHI said:

Homework Statement


For each of the following functions f(z), find f'(z) and identify the maximal region for which f(z) is analytic.

1. f(z)=1/(z^2+1)
2. f(z)=e^{-1/z}

Homework Equations





The Attempt at a Solution


1. f&#039;(z)=\frac{-2z}{(z^2+1)^2} <--this part is easy. I'm having difficulty being certain of the maximum region for analyticity. Here is my attempt.

f(z) is analytic everywhere but + or - i because f'(z) is undefined there.

Is that a true stament or is the correct statement ... f(z) is analytic everywhere but + or - i because f(z) is undefined there.
The first statement is the one you want. There's a theorem that says if a function is complex differentiable at a point, it's analytic at that point. Just because a function exists at a point doesn't mean it's analytic there.
 
vela said:
The first statement is the one you want. There's a theorem that says if a function is complex differentiable at a point, it's analytic at that point.

I don't think so huh vela? Doesn't it have to be complex-differentiable in some disc centered at the point in order for it to be analytic at that point?
 
Yeah, you're right.
 

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