Laurent Series Expansion Centered on z=1

In summary, Laurent series of f(z) about the point z=1 in the annular domain \frac{1}{2}<|z-1|<2 can be found by partial fraction decomposition, as long as \frac{1}{2}<|z-1|<2. When this condition is met, the expansion of f(z) becomes f(z)=-\frac{1}{5}\sum^{\infty}_{n=0}(z-1)^{n}[(-2)^{n+1}-2^{-n-1}] which is the answer.
  • #1
ChemEng1
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0

Homework Statement


Find a Laurent Series of f(z)=[itex]\frac{1}{(2z-1)(z-3)}[/itex] about the point z=1 in the annular domain [itex]\frac{1}{2}<|z-1|<2[/itex].

Homework Equations



The Attempt at a Solution


By partial fraction decomposition, [itex]f(z)=\frac{1}{(2z-1)(z-3)}=\frac{1}{5}\frac{1}{z-3}-\frac{2}{5}\frac{1}{2z-1}[/itex].

When [itex]\frac{1}{2}<|z-1|<2, \frac{(z-1)}{2}<1[/itex].

Hence: [itex]\frac{1}{z-3}=-\frac{1}{3-z}=-\frac{1}{3-z+1-1}=-\frac{1}{2-(z-1)}=-\frac{1}{2}\frac{1}{1-\frac{(z-1)}{2}}[/itex]

Therefore: [itex]\frac{1}{z-3}=-\sum_{n=0}^{\infty}\frac{(z-1)^{n}}{2^{n+1}}[/itex]

And: [itex]f(z)=-\frac{1}{5}\sum_{n=0}^{\infty}\frac{(z-1)^{n}}{2^{n+1}}-\frac{2}{5(2z-1)}.[/itex]

Is this complete? Is there something I could similarly do with the other term? I've tried but haven't been able to find anything that works.
 
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  • #2
I would try explicitly calculating a few terms of the taylor series of the second term and try seeing a pattern.
 
  • #3
It's this, right? [itex]\frac{1}{2z-1}=\sum^{\infty}_{n=0}(-1)^{n+1}(z-1)^{n}*2^{n}[/itex]

So the expansion would become: [itex]f(z)=-\frac{1}{5}\sum^{\infty}_{n=0}(z-1)^{n}[(-2)^{n+1}-2^{-n-1}][/itex]
 
  • #4
I'm not too sure about the n+1 on top of the -1. I'm also not too sure whether there should be a factorial there somewhere.
 
  • #5
ahsanxr said:
I'm not too sure about the n+1 on top of the -1.
You're right about the -1. It should be raise to n, not n+1. I mistakingly thought the index started at 1 instead of 0.
ahsanxr said:
I'm also not too sure whether there should be a factorial there somewhere.
I'm pretty sure the coefficient of the (z-1) term is a multiple of 2 in this case.
 
  • #6
ChemEng1 said:
I'm pretty sure the coefficient of the (z-1) term is a multiple of 2 in this case.

You're correct, sorry for that. I was forgetting to divide by n! in the formula for a_n of a taylor series.
 
  • #7
ahsanxr said:
You're correct, sorry for that. I was forgetting to divide by n! in the formula for a_n of a taylor series.

No biggie. So is the last expression the answer then?
 
  • #8
ChemEng1 said:
No biggie. So is the last expression the answer then?

I still think there are a couple of sign and index errors. Try going through your algebra again to find them (or to prove me wrong).
 
  • #9
I think I figured the other fraction out:

[itex]\frac{1}{2}<|z−1|<2,\frac{1}{2(z-1)}<1.[/itex]

Hence: [itex]\frac{1}{2z-1}=\frac{1}{2z-1-1+1}=\frac{1}{2(z-1)+1}=\frac{1}{2(z-1)}\frac{1}{1+\frac{1}{2(z-1)}}=\frac{1}{2(z-1)}\frac{1}{1-(-\frac{1}{2(z-1)})}=\frac{1}{2(z-1)}\sum^{\infty}_{n=0}(-1)^{n}[2(z-1)]^{-n}=\sum^{\infty}_{n=0}(-1)^{n}[2(z-1)]^{-n-1}[/itex]

That looks much better. I hope.
 

1. What is a Laurent series expansion centered on z=1?

A Laurent series expansion centered on z=1 is a mathematical representation of a complex function in the form of a power series, where the variable z is centered at the point z=1. It includes both positive and negative powers of (z-1) and can be used to approximate the function near z=1.

2. How is a Laurent series expansion different from a Taylor series expansion?

A Taylor series expansion only includes positive powers of (z-1), while a Laurent series expansion includes both positive and negative powers. This allows a Laurent series to represent functions with singularities or poles at z=1, while a Taylor series cannot.

3. What is the region of convergence for a Laurent series expansion centered on z=1?

The region of convergence for a Laurent series expansion centered on z=1 is the set of all complex numbers z for which the series converges. In this case, the region of convergence is an annulus with inner radius 0 and outer radius ∞, centered at z=1.

4. How is the center of a Laurent series expansion chosen?

The center of a Laurent series expansion is chosen based on the location of singularities or poles of the function. In this case, z=1 is chosen as the center because it is the closest singularity to the origin.

5. Can a Laurent series expansion centered on z=1 be used to approximate the function at points other than z=1?

Yes, a Laurent series expansion centered on z=1 can be used to approximate the function at points within the region of convergence, which is an annulus with inner radius 0 and outer radius ∞, centered at z=1. However, the accuracy of the approximation may decrease as the distance from z=1 increases.

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