Complex simplification example, do not see how it followsby TheFerruccio Tags: complex, simplification 

#1
Jan713, 01:38 PM

P: 196

This is an example in Kreyszig Advanced Engineering Mathematics section 14.1, example 6. There are two things I do not understand about the book's example.
1. The problem statement, all variables and given/known data They are integrating around a loop containing [tex]z_0[/tex] with radius [tex]\rho[/tex] of the complex function [tex]\frac{1}{z^m}[/tex]. Would it be helpful for me to just copy the book's example? It's more my attempt to understand the steps that they are taking that is giving me such a hard time. While the title of the problem says "Integral of [tex]\frac{1}{z^m}[/tex] with Integer Power m" the final contour integral they are taking ends up being [tex]\oint_{c}(zz_0)^m dz[/tex] I do not see how this, at all, relates to the title of the problem. Furthermore, I do not understand this equality: [tex](zz_0)^m=p^m e^{imt}[/tex] I apologize for not following the mandated format for this, but it is not really me trying to solve the problem, so much as it is me trying to understand what the heck the textbook is trying to tell me. 



#2
Jan713, 03:39 PM

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, not p.) How did they get to there? For a conformal function, such as z^{m}, integrating around a closed loop containing no poles gives 0. An integral around an arbitrary loop around a pole therefore depends only on the pole  any loop around it will produce the same integral (as long as no other poles get included). So you can replace the original loop with a circle centred on the pole, if that makes things easier. That said, I can't make much sense of z^{m} in this. I would have expected the original function to be (zz_{0})^{m}. 



#3
Jan713, 04:49 PM

P: 196

They started out as... [itex]z(t)=z_0+\rho\left(\cos{t}+i\sin{t}\right)=z_0+ \rho e^{it}[/itex] And, that makes perfect sense to me. They are saying that the curve is represented as the parameterized function [itex]z(t)[/itex] and they are integrating along a circle around [itex]z_0[/itex] with a radius [itex]\rho[/itex] with the shape of the general circle equation [itex]\cos{t}+i\sin{t}[/itex]. From here, they are using Euler's identity to reduce it to [itex]z_0+\rho e^{it}[/itex]. Great, that makes sense. I am still not quite understanding the specific step where they replaced [itex]\left(zz_0\right)^m[/itex] with [itex]\rho^m e^{imt}[/itex]. Is it because [itex]\left(zz_0\right)^m[/itex], in this case is a fixed radius? Maybe I am starting to understand, but I might also be misunderstanding... 



#4
Jan713, 10:21 PM

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Complex simplification example, do not see how it follows
[itex]z(t)=z_0+ \rho e^{it}[/itex]
[itex]zz_0 = \rho e^{it}[/itex] [itex](zz_0)^m = (\rho e^{it})^m = \rho^m (e^{it})^m = \rho^m e^{itm}[/itex] 



#5
Jan813, 11:29 AM

P: 196

Thanks! Makes perfect sense. That's embarrassing. 


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