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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

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

  • 1
    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!2011-10-07

2 Answers 2