# Homework Help: Laurent Series of Rational Functions

1. Jul 21, 2013

### Tsunoyukami

In general I am rather confused by this type of problem. The textbook has a single example and does not show (m)any of its steps so I'm lost. I have a test this coming Thursday and the following is the only question of this type that the prof. has recommended:

"23. Use equations (12) and (13) to find the Laurent expansions of the following rational functions in powers of z and $\frac{1}{z}$ in the indicated region(s).

a) $f(z) = \frac{z+2}{z^{2}-z-2}$ in $1 < |z| < 2$ and then in $2 < |z| < \infty$" (Complex Variables, 2nd edition by Stephen D. Fisher; pg. 152)

I began this problem by factoring the denominator as follows:

$$f(z) = \frac{z+2}{z^{2}-z-2} = \frac{z+2}{(z-2)(z+1)}$$

This form is convenient because we can see that there are poles at $z = 2$ and $z = -1$ which makes sense as to why we must find different Laurent series in regions bounded by circles of modulus 1 and 2. Just to be sure - I could also be asked to find the Laurent series in the region $0 < |z| < 1$, and this Laurent series would be different from either of the ones that the question is asking about, correct?

Next, I wrote the following:

$$f(z) = \frac{z+2}{z-2}{z+1} = f_{1}(z) + f_{2}(z) = \frac{A}{z-2} + \frac{B}{z+1}$$

I solved this and found $A = \frac{4}{3}$ and $B = {-1}{3}$.

I'm not too sure where I'm supposed to go from here, or how to determine where exactly the Laurent series I find is defined...for example, I tired the following:

Consider:

$$f_{1}(z) = \frac{4}{3} \frac{1}{(z-2)} = \frac{4}{3} \frac{1}{-(2-z)} = \frac{4}{3} \frac{-1}{\frac{1}{2}(1-\frac{z}{2})} = \frac{-8}{3} \frac{1}{1-\frac{z}{2}} = \frac{-8}{3} \frac{1}{1 - w} = \frac{-8}{3} \cdot \sum w^{n} = \frac{-8}{3} \cdot \sum (\frac{z}{2})^{n}$$

And a similar computation for $f_{2}(z)$ - however these led me "nowhere". The result I found was $f_{2}(z) = \frac{-1}{3} \frac{1}{z+1} = \frac{-1}{3} \sum (-z)^{n}$.

The book has the following solution:

"In $1<|z|<2$

$$\sum_{- \infty}^{\infty} a_{n}z^{n}$$

$$a_{n} = \frac{(-1)^{n}}{3}, n = -1, -2, ...$$
$$a_{n} = \frac{-4}{3^{n+2}}, n = 0, 1, 2, ...$$

In $2 < |z| < \infty$,

$$\sum_{1}^{\infty} \frac{a_{k}}{z^{k}}$$

$$a_{k} = \frac{(-1)^{k}}{3} + \frac{4}{3^{k+2}}, k = 1, 2, ...$$

I would greatly appreciate any and all help in solving this problem (and other problems of this type) prior to Thursday evening. I'm really lost when it comes to this type of problem...thanks a lot in advance!

EDIT: I think I may have found an error in what I was doing. I need to be careful when writing the series representation of $\frac{1}{1-z}$ that $|z|<1$ - so maybe that led me to the incorrect power series representations...in that manner, I would need to make sure I choose them the right way for the Laurent series to converge within a certain domain...

I'll look more into that later when I have some time. Once again, any help or guidance is appreciated. Thanks!

Last edited: Jul 21, 2013
2. Jul 21, 2013

### HallsofIvy

This is incorrect. What you have written has the "z+1" factor multiplying the fraction and is equivalent to
$$\frac{(z+2)(z+1)}{z-2}$$
What you should have is
$$\frac{z+2}{(z- 2)(z+1)}$$

but apparently that is just a typo since you use it correctly below.

Yes, a Laurent series in the region $0\ge |z|< 1$ would be in powers of z (and, since the function is analytic at z= 0 would be a Taylor's series). A Laurent series in the region $1< |z|< 2$, which is the same as $0< |z- 1|< 1$ would be in powers of z- 1.

3. Jul 21, 2013

### Tsunoyukami

Thanks for catching that typo. It has been corrected in the original post to avoid confusion. Any further thoughts on an approach to solving this type of problem are greatly appreciated. Thanks!

4. Jul 21, 2013

### jackmell

Ok, that's because you're not approaching it from the general case. It takes a little more work that way, but once you master it, you're never ever gonna' have problems with this kind of problem again. I'll start it if you're interested:

Consider $$f(z)=\frac{g(z)}{h(z)}$$

and by partial fractions let's say we can get it to the form:

$$f(z)=\frac{a}{z-p_1}+\frac{b}{z-p_2}+\cdots+\frac{n}{z-p_n}$$

and for illustration purposes, suppose I have the qualitatively similar case:

$$f(z)=\frac{a}{z-2}+\frac{b}{z-3}+\cdots+\frac{n}{z-n}$$

and we which to compute the expansions for $2<|z|<3$. Well surely all the terms with $|p_i|\geq 3$ will have ordinary Taylor series expansions that converge inside the smaller radius of $|z|=3$ right? For example, let's take the p=5 term:

$$\frac{k}{z-5}=\frac{-k}{5(1-\frac{z}{5})}=-\frac{k}{5}\sum_{n=0}^{\infty}\frac{z^n}{5^n}$$

and that series converges for $|z|<5$ so surely it converges for $|z|<2$. I'm just using the relationship

$$\frac{1}{1-u}=\sum_{n=0}^{\infty} u^n,\quad |u|<1$$

We can do this sort of thing for all the terms which have a pole with absolute value greater than the upper bound of the annular domain. However, for the terms less than or equal to the lower bound, we need to compute singular expansions, that is, terms in powers of $1/z$. Take for example our power expansion for $\frac{a}{z-2}$ in the annular domain $2<|z|<3$:

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

and that series will converge for $z>2$ right? So the trick is to convert all the expressions with poles larger than the upper bound, $u$, of the annular domain to a form of $\frac{z}{u}$, and all the expressions with poles less than or equal to the lower bound $v$, to a form $\frac{v}{z}$.

5. Jul 21, 2013

### jackmell

I don't think that's correct Hall as per my previous post.

Last edited: Jul 21, 2013
6. Jul 21, 2013

### Tsunoyukami

I think I understand, at least a bit more...but just to be sure, let me try understanding for my specific function:

$f(z) = \frac{4}{3} \frac{1}{z-2} + \frac{-1}{3} \frac{1}{z+1}$, in the region $1 < |z| < 2$

My function has poles at 1 and 2, the boundaries of the annulus. For the second term of my function, $f_{2}(z) = \frac{-1}{3} \frac{1}{z+1}$ I should write in the form $\frac{v}{z}$ since it is equal to the lower bound. However, the first term of my series has a pole at $z=2$, the upper bound of the annulus - in your description you mention only those poles that are larger than this upper bound. What should I do with this pole? Does it not contribute to the Laurent series in this region then?

By that same logic, if I were to be asked to find the Laurent series in the region $2 < |z| < \infty$ I would have to consider series representations of both terms because both 1 and 2 are less than or equal to the lower bound.

As you can tell, I'm still not 100% on how to do this...

7. Jul 22, 2013

### jackmell

First and importantly, we have to get one right ok? To give us confidence in what we're doing I mean. So for:

$$f(z)=4/3 \frac{1}{z-2}-1/3 \frac{1}{z+1},\quad 1<|z|<2$$

we should get the expansion and then check it numerically with say 50 or more terms. If the agreement is good, we know it's correct. So one of the terms is less than or equal to the lower bound of 1 so that one will be in expansions of 1/z and one of the terms is greater than or equal to the upper bound so that one will be in powers of z. Thusly:

$$\frac{1}{z+1}=\frac{1}{z(1+1/z)}=1/z\sum \frac{(-1)^n}{z^n}$$

$$\frac{1}{z-2}=\frac{1}{-2+z}=-1/2 \frac{1}{1-z/2}=-1/2 \sum \frac{z^n}{2^n}$$

so that:

$$f(z)=4/3\left(1/z\sum \frac{(-1)^n}{z^n}\right)-1/3\left(-1/2 \sum \frac{z^n}{2^n}\right)$$

now before we go to the next one, is that one right? You know how to code that in say Mathematica or whatever to check that?

8. Jul 22, 2013

### Tsunoyukami

I worked through what you have above but have come to a different answer. I believe you have 'mixed up' your coefficients for each term in your final expression for $f(z)$ (I am not sure whether this was a mistake or done on purpose for pedagogical reasons). Below is the work I have done:

$$f(z) = \frac{4}{3} \frac{1}{z-2} - \frac{1}{3} \frac{1}{1+z}$$

We want to write a familiar power series for $\frac{1}{z-2}$ in $z$ because this is the upper bound of the region we are considering.

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

Similarly, we want to write a familiar power series for $\frac{1}{1+z}$ in $\frac{1}{z}$ because this is the lower bound of of the region we are considering. Following the same procedure above I find:

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

Therefore, we can write:

$$f(z) = \frac{4}{3} \left( \frac{-1}{2} \sum_{n = 0}^{\infty} \frac{z^{n}}{2^{n}} \right) - \frac{1}{3} \left( \frac{1}{z} \sum_{n = 0}^{\infty} \frac{(-1)^{n}}{z^{n}} \right)$$

You mentioned that I should check that this is correct by comparing this series representation with the original function. Just to be sure - but this series representation (what I believe is the Laurent series, unless I am mistaken) is valid only in the region $1 < |z| <2$. Therefore any test values I use must be for |z| in this domain, correct? Also, I am not sure how to code this into mathematica or any other useful program to check - nor will I be able to on any given test, so is there a way to be sure without the use of such software?

Next, I have a conceptual question for this type of problem. If I were asked to find the Laurent series in the region $0 < |z| < 1$ I would do a similar calculation as above (except both terms would be greater than or equal to the upper boundary, so both would be expanded in z). Then the series I find would be a Laurent series that converges only in this region. Similarly, what I have found above converges only in $1 < |z| < 2$, correct?

9. Jul 22, 2013

### jackmell

You need to write:

$$f(z) = \frac{4}{3} \left( \frac{-1}{2} \sum_{n = 0}^{\infty} \frac{z^{n}}{2^{n}} \right) - \frac{1}{3} \left( \frac{1}{z} \sum_{n = 0}^{\infty} \frac{(-1)^{n}}{z^{n}} \right),\quad 1<|z|<2$$

pedagogical reasons? No. I just made a mistake.

As far as checking it, easy to do in Mathematica with 25 terms:

Code (Text):

myf[z_] := 4/(3*(z - 2)) - 1/(3*(1 + z));
myseries[z_] := (-2/3)*Sum[z^n/2^n, {n, 0, 25}] - (1/(3*z))*Sum[(-1)^n/z^n, {n, 0, 25}];
point1 = 3/2 + I/4;
N[myf[point1]]
N[myseries[point1]]

Out[7]=
-2.265346534653465 - 1.0534653465346535*I

Out[8]=
-2.2652605956841554 - 1.0553832511573191*I

That's pretty close right? At 50, it's accurate to 4 decimal places. At 100, 11 decimal places. Try and find a machine running Mathematica and check it with 100 terms to get practice.

Correct.

10. Jul 22, 2013

### Tsunoyukami

Thank you very much!

Now that I've made it this far I hope you won't mind checking my work for the region $2 < |z| < \infty$.

We begin with the same function $f(z)$ as in the original post and factor it and find coefficients etc. so that we have $$f(z) = \frac{4}{3} \frac{1}{z-2} + \frac{-1}{3} \frac{1}{z+1}$$. We want to find a Laurent series for the region $2 < |z| < \infty$.

However, both poles are less than or equal to the lower bound and so we expand both $\frac{1}{z-2}$ and $\frac{1}{z+1}$ in $\frac{1}{z}$. For $\frac{1}{1+z}$ this is exactly what I found previously; and converges for $1 < |z|$: $$\frac{1}{1+z} = \frac{1}{z} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{z^{n}}$$.

Similarly, $$\frac{1}{z-2} = \frac{1}{z} \frac{1}{1 - \frac{2}{z}} = \frac{1}{z} \sum_{n=0}^{\infty} \frac{2^{n}}{z^{n}}$$. This series converges for $2 < |z|$.

I can then write the following expression for $f(z)$:

$$f(z) = \frac{4}{3} \frac{1}{z} \sum_{n=0}^{\infty} \frac{2^{n}}{z^{n}} + \frac{-1}{3} \frac{1}{z} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{z^{n}}, 2 < |z| < \infty$$

$$f(z) = \frac{4}{3} \sum_{n=0}^{\infty} \frac{2^{n}}{z^{n+1}} + \frac{1}{3} \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{z^{n+1}}, 2 < |z| < \infty$$

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

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

$$f(z) = \sum_{n=0}^{\infty} \left[ \frac{2^{n+2}}{3} +\frac{(-1)^{n+1}}{3} \right] \frac{1}{z^{n+1}}, 2 < |z| < \infty$$

11. Jul 22, 2013

### jackmell

Outstanding! I would go as far as:

$$f(z) = \frac{4}{3} \sum_{n=0}^{\infty} \frac{2^{n}}{z^{n+1}} + \frac{1}{3} \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{z^{n+1}}, \quad 2 < |z| < \infty$$

Much better results at 25 terms:

Code (Text):

In[1]:=
myseries[z_] := (4/3)*Sum[2^n/z^(n + 1), {n, 0, nMax}] + (1/3)*Sum[(-1)^(n + 1)/z^(n + 1), {n, 0, nMax}];
myf[z_] := 4/(3*(z - 2)) - 1/(3*(z + 1));
nMax = 25;
point = 12.5 - 13.25*I;
N[myseries[point]]
N[myf[point]]

Out[5]=
0.03640674282407563 + 0.04946856286710396*I

Out[6]=
0.03640674282407562 + 0.049468562867103945*I

probably because it's a large number and all the powers are negative which means it converges quickly.