1. ## Laurent series

This is another past exam question, which we have not been given anwers to, so i was hoping someone could check my answer, so i know if i am on the right track.

Find the first three non zero terms of the laurent series on an annulus centred at zero.
f(z)=(1/(z²(1-z)²))

Now the sinularities are at 0, and 1, and they are both double poles.
I did it as follows:

0<|z|<1

(1/(z²(1+(-z)²))=(1/(z²))*∑{n=0}to{∞} zⁿ]
and i used the binomial theorm to get the cooeficients of z.

This gave me:
(1/(z²))-(2/z)+3-4z

Does any one know a way that i can check my solution, to make sure i have not stuffed it up.

2. Originally Posted by thespian
This is another past exam question, which we have not been given anwers to, so i was hoping someone could check my answer, so i know if i am on the right track.

Find the first three non zero terms of the laurent series on an annulus centred at zero.
f(z)=(1/(z²(1-z)²))

Now the sinularities are at 0, and 1, and they are both double poles.
I did it as follows:

0<|z|<1

(1/(z²(1+(-z)²))=(1/(z²))*∑{n=0}to{∞} zⁿ]

I don't understand what you did here: certainly $\frac{1}{z^2(1-z)^2}\neq \frac{1}{z^2(1+(-z)^2)}...!$

So you can do $\frac{\frac{1}{z^2}}{(1-z)^2}=\frac{1}{z^2}\left(1+z+z^2+...\right)^2 =\frac{1}{z^2}(1+2z+3z^2+4z^3+...)=\frac{1}{z^2}+\ frac{2}{z}+3+4z+...$

Tonio

and i used the binomial theorm to get the cooeficients of z.

This gave me:
(1/(z²))-(2/z)+3-4z

Does any one know a way that i can check my solution, to make sure i have not stuffed it up.
.

3. ## Laurent series still need some help

I thought that the series would be alternating from positive to negative,
because the power is -2.
That is :
(1/z^2)-(2/z)+3-4z
Can anyone confirm this because now i am confused.

4. Originally Posted by thespian
I thought that the series would be alternating from positive to negative,
because the power is -2.
That is :
(1/z^2)-(2/z)+3-4z
Can anyone confirm this because now i am confused.

What power are you talking about? The power series is $\frac{1}{1-z}=1+z+z^2+...+z^n+...$ , as can readily be verified using the sum formula for geometric series.
This above, squared, is what you have in the denominator multiplying z^2, you don't have anywhere $\frac{1}{1-z^2}$
You also have to realize that $1+(-z)^2=1+z^2$, so to add that minus sign is pointless.

Tonio

5. (1-z)^-2 doesnt the power series for this alternate between positive and negative?
using the binomial theorm gives me:

1-2z+3z^2-4z^3

because the equation is (1/(z²))((1/((1-z)²)

i am trying to figure out what happens to the exponant of -2 of (1-z)⁻²
because when i expand it using the binomial formula i get alternating sighns.
but you have only positive.

6. let

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

This is the definition of the geometric series

Now taking the derivative of g we get

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

You are trying to find the series for the function

$f(z)=\frac{1}{z^2(1-z)^2}=\frac{1}{z^2}\cdot \frac{1}{(1-z)^2}=\frac{1}{z^2}\cdot g'(z)=...$

7. Originally Posted by thespian
(1-z)^-2 doesnt the power series for this alternate between positive and negative?
using the binomial theorm gives me:

1-2z+3z^2-4z^3

because the equation is (1/(z²))((1/((1-z)²)

i am trying to figure out what happens to the exponant of -2 of (1-z)⁻²

You baffle me: what negative 2 exponent?? It is $\frac{1}{(1-z)^2}=(1-z)^{-2}$ , but the exponent in the denominator is positive, not negative...or at least that's what you wrote already 2-3 times. And you want the positive exponent in the denominator, because if that were a negative one then you'd simply get $\frac{1}{(1-z)^{-2}}=(1-z)^2$ , and this is a simple polynomial.

Tonio

because when i expand it using the binomial formula i get alternating sighns.
but you have only positive.
.