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Complex simplification example, do not see how it follows

by TheFerruccio
Tags: complex, simplification
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TheFerruccio
#1
Jan7-13, 01:38 PM
P: 209
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}(z-z_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](z-z_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.
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haruspex
#2
Jan7-13, 03:39 PM
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Quote Quote by TheFerruccio View Post
They are integrating around a loop containing [itex]z_0[/itex] with radius [itex]\rho[/itex] of the complex function [itex]\frac{1}{z^m}[/itex].
:
I do not understand this equality:
[itex](z-z_0)^m=p^m e^{imt}[/itex]
At this point, it's not just any old loop containing z0, but rather a circle radius ρ centred on z0. So z-z0 will have modulus ρ. (ρ
, not p.) How did they get to there? For a conformal function, such as zm, 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 (z-z0)-m.
TheFerruccio
#3
Jan7-13, 04:49 PM
P: 209
Quote Quote by haruspex View Post
At this point, it's not just any old loop containing z0, but rather a circle radius ρ centred on z0. So z-z0 will have modulus ρ. (ρ
, not p.) How did they get to there? For a conformal function, such as zm, 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 (z-z0)-m.
So, for the equation [itex](z-z_0)^m[/itex] I would understand if it equaled [itex]\rho^m[/itex] Though, I do not quite understand where they got the [itex]e^{imt}[/itex]

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(z-z_0\right)^m[/itex] with [itex]\rho^m e^{imt}[/itex]. Is it because [itex]\left(z-z_0\right)^m[/itex], in this case is a fixed radius? Maybe I am starting to understand, but I might also be misunderstanding...

haruspex
#4
Jan7-13, 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]z-z_0 = \rho e^{it}[/itex]
[itex](z-z_0)^m = (\rho e^{it})^m = \rho^m (e^{it})^m = \rho^m e^{itm}[/itex]
TheFerruccio
#5
Jan8-13, 11:29 AM
P: 209
Quote Quote by haruspex View Post
[itex]z(t)=z_0+ \rho e^{it}[/itex]
[itex]z-z_0 = \rho e^{it}[/itex]
[itex](z-z_0)^m = (\rho e^{it})^m = \rho^m (e^{it})^m = \rho^m e^{itm}[/itex]
Oh, wow, that was dreadfully simple. How did I not see that? :P

Thanks! Makes perfect sense. That's embarrassing.


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