# Homework Help: Laurent series of rational function in annulus

1. Feb 23, 2016

### Incand

1. The problem statement, all variables and given/known data
Find the Laurent expansions of
$f(z) = \frac{z+2}{z^2-z-2}$ in $1 < |z|<2$ and then in $2 < |z|< \infty$
in powers of $z$ and $1/z$.

2. Relevant equations
Theorem:
Let $f$ be a rational function all of whose poles $z_1,\dots , z_N$ in the plane have order one and which has no pole at the origin and which is zero at $\infty$. Suppose that no pole of $f$ lies in the annular region $r<|z|<R$. Then
$f(z) = \sum_{-\infty}^\infty a_kz^k \; \; \; \text{ for } \; \; r < |z|<R,$
where
$a_k=\begin{cases} \sum_{|z_j|< r} z_j^{k-1} \text{Res}(f;z_j), \; \; \; k \le -1\\ -\sum_{|z_j|>R} z_j^{-k-1} \text{Res}(f;z_j), \; \; \; k \ge 0. \end{cases}$

3. The attempt at a solution
We have the residues
$\text{Res}(f;-1) = -\frac{1}{3}$
$\text{Res}(f;2) = \frac{4}{3}$.
In the first case we have $r=1$ and $R=2$. The coefficients are then
$a_k = (-1)^{k-1}\frac{-1}{3} = \frac{(-1)^k}{3}$ for $k=-1,-2,\dots$
$a_k = -2^{-k-1} \frac{4}{3} = \frac{-2}{3\cdot 2^k}$ for $k=0,1,2,\dots$.
The answer however says the first part is correct but the second one should be
$a_k = \frac{-4}{3^{k+2}}, \; \; \; k=0,1,2,\dots$ which makes no sense to me.

For the second region we have $r=2$ and $R=\infty$
Then
$a_k = (-1)^{k-1}\frac{-1}{3} + 2^{k-1}\frac{4}{3}$ for $k=-1,-2,\dots$
If we instead of summing over negative $k$ we write the series as
$\sum_{1}^\infty \frac{a_k}{z_k}$ we have
$a_k = \frac{(-1)^k}{3} +\frac{2}{3\cdot 2^{k}}$.
The answer here however should be
$\sum_{1}^\infty \frac{a_k}{z_k}$ with $a_k=\frac{(-1)^k}{3}+\frac{4}{3^{k+2}}, \; \; \; k=1,2,\dots$.

I'm also wondering about the $|z_j|<r$ and $|z_j|>R$ parts in the theorem, why do I count singularities at $r$ and $R$ when it says strictly less/greater.

I managed to find an old thread covering the exact same question where they seem to get an answer similar to mine: https://www.physicsforums.com/threads/laurent-series-of-rational-functions.702294/

He seem to reach a answer similar to me so perhaps the answer is wrong. However he does it with partial fractions instead of the theorem (although how I understand it the theorem is essential partial fractions). So what I'm looking for is input if I solved this correctly or If I made a mistake? I also seem to get similar errors on other exercises on the method which makes me doubt.

2. Feb 24, 2016

### Samy_A

There seems to be an error in Fisher's book. If you look at how he derives the theorem, it should be:

Let $f$ be a rational function all of whose poles $z_1,\dots , z_N$ in the plane have order one and which has no pole at the origin and which is zero at $\infty$. Suppose that no pole of $f$ lies in the annular region $r<|z|<R$. Then
$f(z) = \sum_{-\infty}^\infty a_kz^k \; \; \; \text{ for } \; \; r < |z|<R,$
where
$a_k=\begin{cases} \sum_{|z_j|< r} z_j^{-k-1} \text{Res}(f;z_j), \; \; \; k \le -1\\ -\sum_{|z_j|>R} z_j^{-k-1} \text{Res}(f;z_j), \; \; \; k \ge 0. \end{cases}$

I checked it for the case $|z|>2$, and then the formula leads to a correct result, same as in the other thread (not the result given in the book, though).

3. Feb 24, 2016

### Incand

Thanks for taking time looking over the theorem and question! It is only a minussign that is missing? It seems to make more sense without it. The example 12c) in the book would also be wrong then since he seem to use the formula there (I assume you have access to the book since you looked up the theorem & proof).

Using the new formula I get
$f(z) = \frac{-2}{3} \sum_0^\infty \frac{z^k}{2^k} + \frac{1}{3}\sum_{1}^\infty \frac{(-1)^k}{z^k}$ for $1 <|z|<2$ and
$f(z) = \frac{1}{3} \sum_1^\infty \frac{(-1) ^k + 2^{k+1}}{z^k}$ for $2<|z|$ which is the same as in the thread (they only shifted the series starting at $0$ instead).

I also tried to to do the same with partial fraction hoping it would illuminate what's going in the theorem (and proof).
$f(z) = \frac{-1}{3(z+1)}+\frac{4}{3(z-2)} = \frac{1}{3} \frac{1}{1+z}-\frac{2}{3} \frac{1}{1-z/2}$. (Partial fraction with residue is really neat!)
In the region $1<|z|<2$ the second half of the function is the geometric series which is convergent in the region
$-\frac{2}{3} \sum_0^\infty (z/2)^n$
However I'm not sure how to get the principal part into $1/z$ form.It seems I could do taylor expansion of $\frac{1}{1+z}$ and then replace the old $z$ with $1/z$ and get $\frac{1}{3} \sum_1^\infty \frac{(-1)^n}{z^n}$ but I have no idea what I just did except that it seems to give the right answer.

I know the same solution is in post #8 but I don't understand that part (so far)

4. Feb 24, 2016

### Samy_A

Yes, just a sign error. And I think that the example is wrong too.

Look at the relevant part of the proof:

In the terms with $z$ in the denominator, the pole is in the numerator, with an exponent one less than the exponent of $z$ in the denominator.
That's not what equation (12) says, as it puts both $z$ and the pole in the denominator.
These look correct.
Actually, the proof in the book looks correct. It's the statement of equation (12) and the example that are wrong. Maybe a typo in (12), and then a blind application of that wrong equation.

5. Feb 24, 2016

### Incand

You're right, I guess it's just a type in the initial formula. I should be more careful reading the proofs in the future instead of skimming them over. Thanks a lot for pointing it out and really going out of your way to help!
I understand the $\frac{1}{z+1} = \frac{1}{z} \frac{1}{1+1/z} =\frac{1}{z} \sum_0^\infty \frac{(-1)^n}{z^n}$ part now as well.

I also seem to get errors on 23b) and c) so I guess there is errors in the answer there as well.

b) $f(z) = \frac{z^2+2z-4}{(z^2-9)(z+1)}$ have residues
$\text{Res}(f;3) = \frac{11}{24}$
$\text{Res}(f;-3) = \frac{-1}{12}$
$\text{Res}(f;-1) = \frac{5}{8}.$
Then by the formula we have in the region $0 \le |z| < 1$ that
$a_k = -\frac{5}{8}(-1)^{-k-1}+\frac{1}{12}(-3)^{-k-1} -\frac{11}{24}3^{-k-1} = \frac{1}{24}[ (-1)^k-2\cdot (-3)^{-k}-11\cdot 3^{-k-1}]$ for $k=0,1,2,\dots$.
And in $1<|z|<3$
$a_k= \begin{cases} \frac{5}{8}(-1)^{k+1}, \; \; \; k=-1,-2,\dots\\ \frac{1}{12}(-3)^{-k-1} -\frac{11}{24}3^{-k-1} , \; \; \; k=0,1,2,\dots \end{cases}$
The second one being equal to $\frac{3}{8}(2\cdot 3^{-k+1}\left((-1)^k-11\right)$

c) I believe this one correct except the 4 in the answer should be a 2.

6. Feb 24, 2016

### Samy_A

I checked this one. The first expression seems correct. I'm not sure about $\frac{1}{24}[ (-1)^k-2\cdot (-3)^{-k}-11\cdot 3^{-k-1}]$.

7. Feb 24, 2016

### Incand

Thanks. It really helps to know if I did it right since I'm self-studing the material, I'm relying on the answers (that turned out to be wrong in this case).

Actually I see lost a factor of $15$ in the last expression. I tried it to write it in the same form as the answer to highlight the difference.
Should be $a_k = -\frac{5}{8}(-1)^{-k-1} + \frac{1}{12}(-3)^{-k-1}-\frac{11}{24}3^{-k-1} = \frac{1}{24}[15(-1)^k+2(-3)^{-k-1}-11\cdot (3)^{-k-1}]$

8. Feb 24, 2016

### Samy_A

Yes, this last one is also correct.
It's annoying that books contain errors, not much you can do about that.
To check a Laurent series, you could do as they did in the other thread, write a program computing the (say) 20 first terms of the series. Comparing the result with the value of the function for a few appropriate values will give a very good indication whether the Laurent series is correct or not.

Last edited: Feb 24, 2016
9. Feb 24, 2016

### Incand

Yea, still It's the first serious errors I found in the book so it's not that bad.
I keep that In mind for next time, I haven't used Mathematica but I could do it just fine In Matlab I guess.