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A problem in my book asks:

In the Laurent series for $\displaystyle f(z) = \frac{1}{(z-4)}$ centered at $z=1$, what is the coefficient of $(z-1)^{-2}$?

The book's solution gives

$$\frac{1}{4-z} = \frac{1}{z-1-3} = \frac{\frac{1}{z-1}}{(1-\frac{3}{(z-1)})} = \frac{1}{z-1} \sum_{n=0}^\infty (\frac{3}{z-1})^n,$$

so the cofficient will be $3$.

However, I don't think this is correct (I don't know complex variables that well, so I'm hesitant to say the book is wrong). First, $f(z)$ is analytic at $z=1$, so shouldn't we get a unique power series expansion about $1$, and hence no negative coefficients?

Second, I'm not sure their expansion of $\displaystyle 1/(1-\frac{3}{(z-1)})$ is valid near $1$, since $\displaystyle |\frac{3}{(z-1)}|$ is not less than $1$ for $z$ close to $1$.

Can anyone confirm my reasoning, or explain why it's wrong?

The way I did this problem was:

$$\frac{1}{z-4} = \frac{1}{(z-1)-3} = \frac{-1}{3-(z-1)} = \frac{-1/3}{1-(z-1)/3}$$

and so we get

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

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I confirm your reasoning, your objections and your final calculation. Everything you wrote is right and the book is wrong.I wish I could upvote you more for your lucidity! – Georges Elencwajg Oct 7 '11 at 7:46

It said "centered at 1"; it didn't say "near 1". There is a Laurent series, centered at 1, valid in $|z-1|<3$. But there is a separate Laurent series, centered at 1, valid in $|z-1|>3$. The first of these, is, indeed, a Taylor series. But the second of these is the one in the book's solution.

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But then what's wrong with this: with the book's expansion, wouldn't we get infinitely many terms of negative degree in a Laurent series centered at 1, implying that $z=1$ is an essential singularity? – user98172389 Oct 7 '11 at 0:44
@user98172389 No. You can deduce properties of a functions only from a series that sums up to the said function (or represents the function) in a neighborhood of the point of interest. This Laurent series only represents the function $f(z)$ far away from the point $z=1$, so you cannot use this series to make deductions about the behavior of $f(z)$ near the point $z=1$. – Jyrki Lahtonen Oct 7 '11 at 3:34

You are right and the book is wrong, or worse than that, is deliberately setting a trap to the student. Here is why.

If you want to study a meromorphic function $f$ near $z=z_0$, you often use its canonical development $f(z)= \sum a_n(z-z_0)^n$ in a Laurent series at $z_0$, with only finitely many negative $n$ for which $a_n\neq 0$. This will tell you the most important things about the germ of $f$ , namely whether the singularity at $z=z_0$ is removable ($\iff a_n=0 \text { for n}\lt 0$) and else the value of its residue at $z_0$, namely $a_{-1}$ .
From the "solution" in your text one would conclude (as you correctly did ) that $f$ has an essential singularity and that the residue is $1$, which are both ridiculous.

Now it is true that if you fix an annulus $A=Ann (z_0; r,R)$ around $z_0$ and consider a holomorphic finction $f\in \mathcal O(A)$ on $A$, you will obtain from these data a Laurent series $L(f;A)=\sum \limits_ {n=-\infty}^{\infty} a_n(z-z_0)^n$, where equality is in the sense of compact (and even normal) convergence on $A$.
But then you must say in advance what the annulus $A$ is. The "solution" in the book would only be correct if you took $A=Ann (1; r,R)$ with $3\lt r \lt R$ which is artificial and most importantly should be stated explicitly in the question.

By the way, I heartily congratulate you on your critical reading of the book and the accurateness of your remarks and objections : this is exactly the right way to study mathematics.

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Before saying "the book is wrong" I would need to see more context. – GEdgar Oct 7 '11 at 14:42
Dear GEdgar, yes, by all means see more context, and if you then think the book is right we'll have an interesting discussion. Meanwhile, I have no reason to assume that the OP has suppressed some significant context to the question. – Georges Elencwajg Oct 8 '11 at 7:32
Unfortunately, the identity of the book is secret... – GEdgar Oct 8 '11 at 12:13
Yes,I was a bit frustrated by that too... – Georges Elencwajg Oct 8 '11 at 17:45

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