SOLVED Laurent series of a function

Apr 2008
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Teyateyaneng
Calculate the Laurent development of the function \(\displaystyle f(z)=\frac{1}{(z-a)(z-b)}\) in the region \(\displaystyle \Omega: 0<|a|<|z|<|b|<+\infty\).
My attempt: I notice that f has 2 singularities, one in z=a, the other in z=b. By intuition, the radius of convergence of the series should be \(\displaystyle \frac{|b|-|a|}{2}\) but it's not asked.
Ok so my problem is where do I center the Laurent series? There's the possibility that f is analytic in \(\displaystyle \Omega\) (and so the Laurent series is worth the Taylor's series) but I'm unsure since it doesn't seem bounded when \(\displaystyle z\to a^+\) and \(\displaystyle z\to b^-\) so I'd bet it's not analytic in omega.
Now \(\displaystyle \frac{1}{z-a}=\frac{1}{z} \sum _{n=0}^{+\infty} \left ( \frac{a}{z} \right ) ^n\) with \(\displaystyle \left | \frac{a}{z} \right |<1 \Rightarrow |a|<|z|\) which is ok in the region omega. I could get a similar result for \(\displaystyle \frac{1}{z-b}\) but it wouldn't be valid in omega so I don't really know how to go further.
So I can write \(\displaystyle f(z)=\frac{1}{z(z-b)} \sum _{n=0}^{+\infty} \left ( \frac{a}{z} \right ) ^n\) and I'm stuck.
 

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\(\displaystyle \displaystyle{\frac{1}{(z-a)(z-b)}=\frac{1}{(b-a)(z-b)}+\frac{1}{(a-b)(z-a)}} \)
 
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Apr 2008
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Teyateyaneng
Ok thanks for the help. I still have the same problem. While I can get an expression for \(\displaystyle \frac{1}{(a-b)(z-a)}\) (it's worth \(\displaystyle \frac{1}{z(a-b)} \sum _{n=0}^{+\infty} \left ( \frac{a}{z} \right ) ^n\).), I get stuck to get an infinite series expression for something with \(\displaystyle \frac{1}{z-b}\) because in omega, \(\displaystyle |z|<b\).
 
Jul 2010
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Since |z|<|b| in omega and |b|> 0 for all complex numbers b, \(\displaystyle |\frac{z}{b}|<1\) in omega.

In this case: \(\displaystyle \frac{1}{z-b} = \frac{1}{b}\frac{1}{z/b-1} = -\frac{1}{b}\frac{1}{1-z/b} = -\frac{1}{b}\sum_{n=0}^{\infty}(\frac{z}{b})^n \)

In the other case we have |z|>|b| and then \(\displaystyle |\frac{b}{z}|<1\).

In this case: \(\displaystyle \frac{1}{z-b} = \frac{1}{z}\frac{1}{1-b/z} = \frac{1}{z}\sum_{n=0}^{\infty}(\frac{b}{z})^n \)
 
Jul 2010
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As for your more general question, here is what I think. Lets talk about your function f(z). It is a quotient of polynomials where the numerator is a constant and the denominator is a product of two linear factors. There is a theorem somewhere saying that such functions are analytic everywhere except at their singularities. In your case z = a or z = b. From this, you can see the f(z) must be analytic in \(\displaystyle \Omega\), which is an annulus.

Now, I'm not 100% sure on this, but talking about radius of convergence for a Laurent series can be a bit of a misnomer. We should really be talking about a region in which the series converges. For similar reasons, the whole issue of where you center the series is a bit superfluous as well, because of the following:

If you center the Laurent series around 0, it'll equal the Taylor series for f(z) since it is analytic on |z| < |a|, and it'll terminate at the nearest singularity, which is a.

Similarly, in omega, you can choose a point and derive a Taylor series about the point, and it'll terminate at whichever singularity is closer a or b. This will depend on the placement of said point in relation to a and b.

Now, the point of the Laurent series in this case would be to find a series representation of f(z) which will be analytic in the whole annulus omega. So, in this case we are not really looking at a point at which we want to center the series. Rather, we are looking at a way to manipulate f(z) to look like a the sum of a geometric series. In this example, it should be pretty easy, because the function is already looking ready.

So you have the first step, which is to split apart f(z). You can actually do without the splitting, but the downside is that you will have to multiply series and this can be messy. So:

\(\displaystyle \frac{1}{(z-a) (z-b)}=\frac{1}{(b-a) (z-b)}+\frac{1}{(a-b) (z-a)} = \frac{1}{b-a}(\frac{1}{z-b}-\frac{1}{z-a})\)

Now you have the following inequalities: |a| < |z| and |z| < |b|. You can see where this is going. From these we need to find expressions of z that have magnitude < 1 to use converging geometric series. From the first inequality we get |a\z| < 1 and from the second |z\b| < 1. Now you are pretty much ready.

\(\displaystyle \frac{1}{z-b} = -\frac{1}{b}\frac{1}{1-z/b} \) and \(\displaystyle \frac{1}{z-a} = \frac{1}{z}\frac{1}{1-a/z} \)

So now \(\displaystyle f(z) = \frac{1}{a-b}(\frac{1}{b}\frac{1}{1-z/b} +\frac{1}{z}\frac{1}{1-a/z})\). From here you can just use geometric series and maybe simplify a bit.
 
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Apr 2008
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Teyateyaneng
Ok thank you very much. I reach \(\displaystyle f(z)=\left ( \frac{1}{a-b} \right ) \left [ \frac{1}{z} \sum _{n=0}^{+\infty} \left ( \frac{a}{z} \right ) ^n - \frac{1}{b} \sum _{n=0}^{+\infty} \left ( \frac{z}{b} \right ) ^n \right ]\) which is still not a Laurent series but I will try to work on it.
I've a question however: how did you know to split f(z) as \(\displaystyle {\frac{1}{(z-a)(z-b)}=\frac{1}{(b-a)(z-b)}+\frac{1}{(a-b)(z-a)}} \)? I wouldn't be able to do it on my own if I don't understand the reason.
 
Jul 2010
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Ok thank you very much. I reach \(\displaystyle f(z)=\left ( \frac{1}{a-b} \right ) \left [ \frac{1}{z} \sum _{n=0}^{+\infty} \left ( \frac{a}{z} \right ) ^n - \frac{1}{b} \sum _{n=0}^{+\infty} \left ( \frac{z}{b} \right ) ^n \right ]\) which is still not a Laurent series but I will try to work on it.
I've a question however: how did you know to split f(z) as \(\displaystyle {\frac{1}{(z-a)(z-b)}=\frac{1}{(b-a)(z-b)}+\frac{1}{(a-b)(z-a)}} \)? I wouldn't be able to do it on my own if I don't understand the reason.
I think there is a + the nor a -, but I'm not 100% sure. Also, it is almost a Laurent series, you just have to do a small transformation on the coefficients in the left sum to make it of the form \(\displaystyle \sum _{n=1}^{+\infty} \frac{c_{n}}{z^n}\)
 

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I've a question however: how did you know to split f(z) as \(\displaystyle {\frac{1}{(z-a)(z-b)}=\frac{1}{(b-a)(z-b)}+\frac{1}{(a-b)(z-a)}} \)? I wouldn't be able to do it on my own if I don't understand the reason.
partial fractions.
 
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Apr 2008
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Teyateyaneng
I think there is a + the nor a -, but I'm not 100% sure.[/tex] Hmm I don't really see what you mean.
\(\displaystyle Also, it is almost a Laurent series, you just have to do a small transformation on the coefficients in the left sum to make it of the form \(\displaystyle \sum _{n=1}^{+\infty} \frac{c_{n}}{z^n}\)\)
\(\displaystyle
For the left sum, I reach that it's worth \(\displaystyle \sum _{n=-\infty}^{-1} h^n z^n\) where \(\displaystyle h^n=\frac{1}{a^{n+1}}\).
I'm not sure I can easily "glue" both infinite series into a Laurent series.
Thanks chiph588, I got it.\)
 
Jul 2010
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\(\displaystyle f(z) = \frac{1}{a-b}(\frac{1}{b}\frac{1}{1-z/b} +\frac{1}{z}\frac{1}{1-a/z}) = \frac{1}{a-b}(\frac{1}{b}\sum_{n=0}^{\infty}(\frac{z}{b})^n+\frac{1}{z}\sum_{n=0}^{\infty}(\frac{a}{z})^n)\)

\(\displaystyle = \frac{1}{a-b}(\sum_{n=0}^{\infty}\frac{z^n}{b^{n+1}}+\sum_{n=0}^{\infty}\frac{a^n}{z^{n+1}}) = \frac{1}{a-b}(\sum_{n=0}^{\infty}\frac{z^n}{b^{n+1}}+\sum_{n=1}^{\infty}\frac{a^{n-1}}{z^n})\)

\(\displaystyle = \frac{1}{a-b}(\sum_{n=0}^{\infty}\frac{z^n}{b^{n+1}}+\sum_{n=-\infty}^{-1}\frac{z^n}{a^{n+1}}) = \sum_{n=-\infty}^{\infty}c_n z^n\)

where \(\displaystyle c_n = \frac{1}{(a-b)a^{n+1}}\) if n smaller or equal to -1
and \(\displaystyle c_n = \frac{1}{(a-b)b^{n+1}}\) if n is greater or equal to 0
 
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