Series of 1/z for 0<|z|<1

**quidh** · September 17th 2011, 08:33 AM

As in title I'm trying to write the Laurent series of $1/z$ for $0<|z|<1$ , but I'm stuck...

**FernandoRevilla** · September 17th 2011, 08:52 AM

Originally Posted by quidh

As in title I'm trying to write the Laurent series of $1/z$ for $0<|z|<1$ , but I'm stuck...

Hint: For all $z\neq 0$ we have

$\dfrac{1}{z}=\ldots +\dfrac{0}{z^3}+\dfrac{0}{z^2}+\dfrac{1}{z}+0+0z+0 z^2+0z^3+\ldots$

**quidh** · September 17th 2011, 09:17 AM

$\sum_{n=0}^{+\infty} z^{n-1}0^n$ ???
If $n=0$ I have $\frac{1}{z}$ , while if $n$ is not $0$ I have $0$ . Right?

Edit: Sorry $0^0$ is indeterminate... I really have no idea...

**FernandoRevilla** · September 17th 2011, 10:55 AM

Don't complicate things (

) the solution is exactly:

$\dfrac{1}{z}=\ldots +\dfrac{0}{z^3}+\dfrac{0}{z^2}+\dfrac{1}{z}+0+0z+0 z^2+0z^3+\ldots\quad (0<|z|<+\infty)$

And of course, the expansion is also valid for $0<|z|<1$ .

**quidh** · September 17th 2011, 11:57 AM

Thanks for your help,
Sorry for my bad english, but probably the one you wrote is the "expansion"; I've to write it in a "series form" ( $\sum_{n=0}^{+\infty}$ )... I'm a bit confused.

**Opalg** · September 17th 2011, 12:28 PM

Originally Posted by quidh

Thanks for your help,
Sorry for my bad english, but probably the one you wrote is the "expansion"; I've to write it in a "series form" ( $\sum^{+\infty}_{\color{red}n=0}$ ). That should be $\color{red}n=-\infty$ , not $\color{red}n=0.$

In a Laurent series, n has to go from $-\infty$ to $\infty,$ not just from 0 to $\infty.$ For the function 1/z, as FernandoRevilla has already pointed out (twice), all the terms in the Laurent series are 0, except when n=–1.

**quidh** · September 17th 2011, 01:12 PM

Ok, another last thing, the exercise was at the beginning: Laurent series of $\frac{z^3+1}{z^2(z-1)}$ for $0<|z|<1$ ,
then I wrote partial fractions: $\frac{z^3+1}{z^2(z-1)} = \frac{z}{z-1}-\frac{1}{z^2}-\frac{1}{z}+\frac{1}{z-1}$ and then I wrote the Laurent Series for every fraction:

$\frac{1}{z-1} = -\sum_{n=0}^{+\infty}z^n$

$\frac{z}{z-1}= -\sum_{n=0}^{+\infty}z^{n+1}$

At this point my question is: I can leave $-\frac{1}{z}$ and $-\frac{1}{z^2}$ as they are?

So the final result will be:

$\frac{z^3+1}{z^2(z-1)} = -\sum_{n=0}^{+\infty}z^n-\sum_{n=0}^{+\infty}z^{n+1} -\frac{1}{z^2}-\frac{1}{z}$

**chisigma** · September 17th 2011, 10:13 PM

Originally Posted by Opalg

In a Laurent series, n has to go from $-\infty$ to $\infty,$ not just from 0 to $\infty.$ For the function 1/z, as FernandoRevilla has already pointed out (twice), all the terms in the Laurent series are 0, except when n=–1.

If it is allowed to consider a Taylor expansion a particular case of Laurent expansion where all the negative coefficients are 0, then the expansion of $f(z)=\frac{1}{z}$ around $z=\frac{1}{2}$ ...

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

... is a Laurent expansion. The (1) converges for $|1-2 z|<1$ but with a procedure called 'analytic extension' the analyticity of $\frac{1}{z}$ can be 'extended' to the region $0<|z|<1$ . Of course a spontaneous question is: what's the practical utility of all that?...

Kind regards

$\chi$ $\sigma$

**Opalg** · September 18th 2011, 02:49 AM

Originally Posted by quidh

Ok, another last thing, the exercise was at the beginning: Laurent series of $\frac{z^3+1}{z^2(z-1)}$ for $0<|z|<1$ ,
then I wrote partial fractions: $\frac{z^3+1}{z^2(z-1)} = \frac{z}{z-1}-\frac{1}{z^2}-\frac{1}{z}+\frac{1}{z-1}$ and then I wrote the Laurent Series for every fraction:

$\frac{1}{z-1} = -\sum_{n=0}^{+\infty}z^n$

$\frac{z}{z-1}= -\sum_{n=0}^{+\infty}z^{n+1}$

At this point my question is: I can leave $-\frac{1}{z}$ and $-\frac{1}{z^2}$ as they are? Yes.

So the final result will be:

$\frac{z^3+1}{z^2(z-1)} = -\sum_{n=0}^{+\infty}z^n-\sum_{n=0}^{+\infty}z^{n+1} -\frac{1}{z^2}-\frac{1}{z}$

The final answer is correct, but it looks a bit clumsy because of the two separate summations, which could easily be combined into one. In fact, it would have been better to write the partial fraction expression using only one term with denominator $z-1:$

$\frac{z}{z-1}+\frac{1}{z-1} = \frac{z+1}{z-1} = \frac{(z-1)+2}{z-1} = 1 + \frac2{z-1}.$

That way, you get a constant term 1 for the Laurent series, plus a single series arising from the z-1 denominator.

**HallsofIvy** · September 18th 2011, 04:06 PM

Originally Posted by quidh

Thanks for your help,
Sorry for my bad english, but probably the one you wrote is the "expansion"; I've to write it in a "series form" ( $\sum_{n=0}^{+\infty}$ )... I'm a bit confused.

'
Yes, you are. A "Laurent series" typically cannot be written in that form- that's what distinguishes it from a Taylor's series. A Laurent series has negative powers. You can write a Laurent series in the form $\sum_{n=-\infty}^\infty a_nz^n$ .

In this case just define $a_{-1}= 1$ , $a_n= 0$ for all other n.

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