# Thread: Laurent series at a point

1. ## Laurent series at a point

I must find the Laurent series of $\displaystyle \frac{1}{2z-i}$ at the point $\displaystyle a=-1$ and find the radius of convergence of the series.
My attempt: I guess they mean $\displaystyle z=-1$ when they say $\displaystyle a=-1$?
First I realize that $\displaystyle f(z)=\frac{1}{2z-i}$ has a singularity at $\displaystyle z=\frac{i}{2}$ and so the radius of convergence of the Laurent series centered at $\displaystyle z=-1$ should be equal to $\displaystyle \frac{\sqrt 5}{2}$.
I'm not sure if I should make a change of variable in order to get the Laurent series centered at $\displaystyle z=-1$ or if I should make any other algebra.
What I've done so far is to write $\displaystyle f(z)$ as $\displaystyle \frac{1}{2z} \left ( \frac{1}{1-\frac{i}{2z}} \right )=\frac{1}{2z} \sum _{n=0}^{+\infty} \left ( \frac{i}{2z} \right ) ^n =\sum _{n=0}^{+\infty} \frac{i^n}{(2z)^{n+1}}$ for $\displaystyle \left | \frac{i}{2z} \right | <1 \Rightarrow \left | z \right | <\frac{1}{2}$. But I'm still far from a series of the form $\displaystyle \sum _{n=-\infty}^{\infty} a_n (z+1)^n$.
Any tip would be good for me.

2. One way is to note that the Laurent series around a=-1 wll be the functions Taylor expansion, so try to derive that one instead.

3. Originally Posted by Jose27
One way is to note that the Laurent series around a=-1 wll be the functions Taylor expansion, so try to derive that one instead.
Ok but how did you notice this? I thought that the function had to be analytic everywhere in its domain for both series to be the same. But since there's a singularity in z=i/2, can you tell me why the Laurent series and Taylor's series will be the same?
And ok, I will find the Taylor series.
Edit: I found $\displaystyle \sum _{n=0}^{+\infty} \frac{2^n (z+1)^n}{\log (-2-i)n!}$.
Edit 2: And so the radius of convergence of the series is $\displaystyle +\infty$, so it converges for all z in $\displaystyle \mathbb{C}$ so I certainly made at least 1 error...

4. $\displaystyle \frac{1}{2z-i}=\frac{1}{2(z+1)-(2+i)} = \frac{-1}{2+i}\frac{1}{1-\frac{2}{2+i}(z+1)} = \frac{-1}{2+i}\sum_{j=0}^\infty \left(\frac{2}{2+i}(z+1)\right)^j$ for $\displaystyle \left|\frac{2}{2+i}(z+1)\right|<1$.

I don't know how you got logs in there!

The Taylor series of a function around a point of analyticity is a specific case of a Laurent series : the Taylor series is the Laurent series there. However, around a singularity, there is no Taylor series, but there may be a Laurent series. In this case, since the function is analytic at $\displaystyle -1$, the Laurent series you are being asked to find is nothing else than the Taylor series there.

5. Originally Posted by Bruno J.
$\displaystyle \frac{1}{2z-i}=\frac{1}{2(z+1)-(2+i)} = \frac{-1}{2+i}\frac{1}{1-\frac{2}{2+i}(z+1)} = \frac{-1}{2+i}\sum_{j=0}^\infty \left(\frac{2}{2+i}(z+1)\right)^j$ for $\displaystyle \left|\frac{2}{2+i}(z+1)\right|<1$.

I don't know how you got logs in there!

The Taylor series of a function around a point of analyticity is a specific case of a Laurent series : the Taylor series is the Laurent series there. However, around a singularity, there is no Taylor series, but there may be a Laurent series. In this case, since the function is analytic at $\displaystyle -1$, the Laurent series you are being asked to find is nothing else than the Taylor series there.
Ok thanks a lot. So in fact they are asking the Laurent series of f at a single point where f is analytic (and f is analytic in its neighborhood) and that's why it coincides with the Taylor series. Is that right?
I also don't know how I got logs, it seems I've integrated something but now I don't understand what I've done. (I just hope it won't happen in the exam!).
Now I get the Taylor series $\displaystyle \sum _{n=0}^{+\infty} \frac{(-1)^n 2^n(z+1)^n}{(-2-i)^{n+1}n!}$ which seems also wrong compared to yours. And I reach a radius of convergence of $\displaystyle +\infty$ which also makes no sense.
Am I right to say that $\displaystyle f^{n}(z)=\frac{(-1)^n 2^n}{(2z-i)^{n+1}}$?
From your condition $\displaystyle \left|\frac{2}{2+i}(z+1)\right|<1$ I get a condition on z: it's a circle centered at (-1,0) and with radius $\displaystyle \frac{\sqrt 5}{2}$ which makes sense to me! See my first post for the radius of convergence.
Another question: how did you know that writing 2z-i as something with "2+i" as factor was a good idea? I don't think it might occur to me to do what you did.

6. Yes, the Taylor series and the Laurent series coincide where the function if they are around a point where f is analytic. The way you would derive the Taylor series is to find an expression of the n-th derivative of f(z), evaluate it at the point of expansion (z = -1 in this case) and then do some simplification.

$\displaystyle f(z) = \frac{1}{2z-i}$

1st derivative$\displaystyle f'(z) = \frac{-2}{(2z-i)^{2}} = \frac{(-2)^{1} 1!}{(2z-i)^{2}}$

2nd derivative $\displaystyle f''(z) = \frac{8}{(2z-i)^{3}} = \frac{(-2)^{2} 2!}{(2z-i)^{3}}$

and in general we get $\displaystyle f^{(n)}(z) = \frac{(-2)^{n} n!}{(2z-i)^{n+1}}$

Those factorials tend to creep up in derivations like this, and looking at the expression you got, the only difference is that you are missing said factorial.

Now the Taylor series is:

$\displaystyle \sum_{n=0}^{\infty}f^{(n)}(-1)\frac{(z+1)^{n}}{n!} = \sum_{n=0}^{\infty}\frac{(-2)^{n}n!}{(2(-1)-i)^{n+1}}\frac{(z+1)^{n}}{n!}$ $\displaystyle = -\sum_{n=0}^{\infty}\frac{2^{n}}{(2+i)^{n+1}}(z+1)^ {n} = \frac{-1}{2+i}\sum_{n=0}^{\infty}(\frac{2}{2+i}(z+1))^{n}$

after canceling out the n powers of -1, the n!, bringing out a 1/(2+i) and collecting the remaining terms under a power of n.

The condition $\displaystyle \left|\frac{2}{2+i}(z+1)\right|<1$ is due to the fact that we have a geometric sum and it only converges if the thing's magnitude under the power of n is smaller than 1.

If you look at Bruno's derivation, he is doing the complex variables derivation, which in this case is easier. You are trying to get an expression that looks as much as a geometric sum as possible. Here is the prototypical example:

$\displaystyle \frac{1}{1-x} = \sum_{n=0}^{\infty}x^{n}$ where we need x < 1

In your case you only have 1/(2z-i). We want to get something of the form 1/(1-x) where x is some expression containing z and is smaller than 1. You can start off by thinking of x = 2z, but the magnitude of 2z > 1, so we cannot do that. So here is how it goes.

First we need to have z+1 and not just z in the expression, since we want the series to be expanded around z = -1. So you add and subtract 1 to get:

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

And then you do the rest.

7. Thank you Vlasev, that was extremely helpful.
Edit: By the way, is the radius of convergence worth $\displaystyle \frac{\sqrt 5}{2}$?