MHB Where Does the Sign of n Matter in Palka's Examples?

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The discussion focuses on the relevance of the sign of n in Bruce P. Palka's examples from "An Introduction to Complex Function Theory." In Example 1.1, the assumption of n being positive is crucial because it ensures that powers of z decrease to zero, while for negative n, the powers do not converge, leading to an infinite series. Similarly, in Example 1.2, the calculations appear valid for both positive and negative n, but the proof is specifically tailored for positive n, necessitating a separate proof for negative n. The complexity of infinite series is highlighted as a reason Palka may have avoided them in his initial examples. Understanding these distinctions is essential for grasping the differentiation rules presented in the text.
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I am reading Bruce P. Palka's book: An Introduction to Complex Function Theory ...

I am focused on Chapter III: Analytic Functions, Section 1.2 Differentiation Rules ...

I need help with some aspects of Examples 1.1 and 1.2, Section 1.2, Chapter III ...

Examples 1.1 and 1.2, Section 1.2, Chapter III read as follows:
View attachment 9334
My questions regarding the above two examples from Palka are as follows:Question 1

Can someone please explain where in the calculations of Example 1.1 does the assumption of n being positive becomes relevant ...

I am puzzled because it appears that each of the steps of the calculation are true whether n is positive or negative ...

Question 2

Can someone please explain where in the calculations of Example 1.2 does the assumption of n being negative becomes relevant ...

I am puzzled because it appears that each of the steps of the argument/calculation are true whether n is positive or negative ...Hope someone can help ...

Help will be much appreciated ...

Peter
 

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In z^n- z_0^n= (z- z_0)(z^{n-1}+ z_0z^{n-2}+ \cdot\cdot\cdot+ z_0^{n-2}z+ z_0^{n-1}), with n positive, powers of z are decreasing and eventually become 0 in the z_0^{n-1} term. If n is negative powers of z decreasing become more negative so do not eventually become 0. We get a infinite series. It might well be true that you could prove the desired statement using infinite series but that Palka wants to avoid the additional complications of infinite series (which may not have been introduced at this point). And since the proof given is only for n positive, a separate proof has to be given for n negative.
 
HallsofIvy said:
In z^n- z_0^n= (z- z_0)(z^{n-1}+ z_0z^{n-2}+ \cdot\cdot\cdot+ z_0^{n-2}z+ z_0^{n-1}), with n positive, powers of z are decreasing and eventually become 0 in the z_0^{n-1} term. If n is negative powers of z decreasing become more negative so do not eventually become 0. We get a infinite series. It might well be true that you could prove the desired statement using infinite series but that Palka wants to avoid the additional complications of infinite series (which may not have been introduced at this point). And since the proof given is only for n positive, a separate proof has to be given for n negative.

Thanks for the help, HallsofIvy ...

Peter
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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