Laurent series representation of f(z)=(z-1)/(z-2) at z=i

In summary, to find the Laurent series representation for f(z)=(z-1)/(z-2) at z=i, we can first find the Taylor series expansion by differentiating the function and evaluating at z=i. This will give us a series with positive powers of (z-i). However, since z=2 is a pole, we can also find the Laurent series expansion by considering the
  • #1
TheCly
2
0

Homework Statement


Find the Laurent series representation for f(z)=(z-1)/(z-2) at z=i.


Homework Equations



NA

The Attempt at a Solution


I have taken multiple derivatives but I keep getting stuck at what to do after I find my representation of my nth derrivative. PLEASE HELP
 
Physics news on Phys.org
  • #2
Why did you take the derivatives? I'm not saying that's not the right thing to do, but you must have had a reason for doing that.
 
  • #3
because to find the Laurent serries representation you must first find the talor expansion
 
  • #4
TheCly said:
because to find the Laurent serries representation you must first find the talor expansion

No. The Taylor expansion is analytic entirely in it's domain of convergence and contains positive powers of (z-z0). The Laurent series is an expansion about a singular point in the domain of convergence and contains negative powers of (z-z0). Now look at your function. At the point i, it is analytic with the nearest singular point at z=2. Therefore, we can represent the function with a Taylor series with a radius of convergence equal to the distance to the nearest singular point which would be z=2. Contrast that with expanding it about the singular point z=2. In that case, the power series would be a Laurent series with negative powers of (z-2). Also, you don't have to compute the derivatives to find this Taylor series. Suppose I had just:

[tex]\frac{1}{z-2}[/tex]

and I wanted to expand it about z=i. I could write:

[tex]\frac{1}{z-2}=\frac{1}{z-i+i-2}=\frac{1}{(i-2)+(z-i)}[/tex]

[tex]=\frac{1}{(i-2)\left[1+\frac{z-i}{i-2}\right]}[/tex]

[tex]=\frac{1}{i-2}\sum_{n=0}^{\infty} \frac{(-1)^n (z-i)^n}{(i-2)^n }[/tex]

Now note that I had just 1 in the numerator. Can you long-divide yours, then do what I did to obtain it's Taylor Series?
 
Last edited:
  • #5
jackmell said:
The Laurent series is an expansion about a singular point in the domain of convergence and contains negative powers of (z-z0).
Not necessarily. You can expand about any point z0. If the function is analytic there, you get no negative powers, and the Laurent series reduces to the Taylor series. In this case, the point z0=i isn't a pole, so TheCly just needs to find the Taylor series, which can be done following the usual method of differentiating f(z) and evaluating

[tex]f(z) = f(z_0)+f'(z_0)(z-z_0)+\frac{f''(z_0)}{2!}(z-z_0)^2+\cdots[/tex]
 
  • #6
Ok. Thanks Vela. I should have stated the Laurent series can contain positive powers of (z-z0) and negative powers but a Taylor series will only contain positive powers of (z-z0). Also, I'd like to suggest to the OP that he check his answer by seeing if it agrees, the first few terms anyway, with the series expansion in Mathematica or Wolfram Alpha.

Just type:

Series[(z-1)/(z-2),{z,I,5}]
 
  • #7
I think some people do say a series isn't a Laurent series unless it has negative powers, but at least the way I learned it (and in a few sources I checked), the Laurent series may but doesn't necessarily have negative powers and, when it doesn't, it reduces to the Taylor series.

Which brings me to my next point. What I wrote earlier wasn't quite correct. The poles of a complex function divide the complex plane into some combination of a circle and one or more annuli centered at z0. The Laurent series reduces to the Taylor series if the region of convergence you're interested in is the circle about z0. If you're looking for the series outside of that circle, you'll get negative powers even if the function is analytic at z0.

In this problem, there's a pole at z=2, which divides the complex plane into two regions centered at z0=i: the first is |z-i|<√5, and the other, |z-i|>√5. So I'm guessing TheCly is supposed to find the series for both regions; in this context, the comment about needing to find the Taylor series first makes a bit more sense.
 
  • #8
vela said:
If you're looking for the series outside of that circle, you'll get negative powers even if the function is analytic at z0.

Yeah, a good exercise is to take the function:

[tex]\frac{1}{(z-z_0)(z-z_1)\cdots(z-z_n)}[/tex]

and calculate the Laurent series for each annuli between the singular points. For the particular problem here, for [itex]|z-i|>\sqrt{5}[/itex], we could just group the terms differently:

[tex]\frac{z-1}{z-2}=1+\frac{1}{(z-i)\left[1+\frac{i-2}{z-i}\right]}[/tex]
 

Related to Laurent series representation of f(z)=(z-1)/(z-2) at z=i

1. What is a Laurent series representation?

A Laurent series representation is a power series expansion of a complex function that allows for both positive and negative powers of the variable z. It is used to approximate functions that are not defined at certain points or have singularities.

2. How is the Laurent series representation of a function calculated?

The Laurent series representation of a function is calculated by first finding the Laurent coefficients, which are the coefficients of the positive and negative powers of z. These coefficients can be found by multiplying the function by z^n and integrating over a closed curve around the point of interest. The resulting series is then the Laurent series representation of the function.

3. What is the purpose of calculating the Laurent series representation?

The Laurent series representation is useful for approximating complex functions and analyzing their behavior near singularities. It can also be used to find the residues of the function, which are important in calculating integrals and solving differential equations.

4. How does the Laurent series representation of f(z)=(z-1)/(z-2) at z=i help in understanding the behavior of the function?

At z=i, the function f(z) has a singularity. The Laurent series representation helps in understanding the nature of this singularity and how the function behaves near it. It also allows for the calculation of the residue, which is needed for evaluating integrals involving this function.

5. Can the Laurent series representation be used to approximate the function f(z)=(z-1)/(z-2) at points other than z=i?

Yes, the Laurent series representation can be used to approximate the function at any point where it is not defined or has a singularity. However, the accuracy of the approximation may vary depending on the distance from the point of interest and the radius of convergence of the series.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
399
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
12
Views
4K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
3K
  • Calculus and Beyond Homework Help
Replies
2
Views
970
  • Calculus and Beyond Homework Help
Replies
6
Views
958
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
17
Views
1K
Back
Top