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

[TheCly]

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

 

[vela]

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.

 

[TheCly]

because to find the Laurent serries representation you must first find the talor expansion

 

[jackmell]

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:

\frac{1}{z-2}

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

\frac{1}{z-2}=\frac{1}{z-i+i-2}=\frac{1}{(i-2)+(z-i)}

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

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

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?

 

[vela]

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

f(z) = f(z_0)+f'(z_0)(z-z_0)+\frac{f''(z_0)}{2!}(z-z_0)^2+\cdots

 

[jackmell]

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}]

 

[vela]

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.

 

[jackmell]

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:

\frac{1}{(z-z_0)(z-z_1)\cdots(z-z_n)}

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

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