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My book claims that $1/(1-z)$ has an essential singularity at $z=1$ by writing out the Laurent series and showing that there are infinitly many terms. Why isn't this just a pole of order 1? We can write this as $-1/(z-1)$, then the numerator is a nonvanishing analytic function, and the denominator is of the form $(z-1)$, so it seems like by definition this should be a pole, and not an essential singularity. What am I missing here?

If anyone is curious: the book is the Princeton Review for the GRE Math Subject Test, 4th edition, p. 312-313

Their argument is this. Consider the region $1 < |z| < \infty$ and write

$f(z) = \frac{1}{1-z} = \frac{1}{z(\frac{1}{z}-1)} = -\frac{1}{z} \cdot \frac{1}{z-\frac{1}{z}}$.

Then if $|z| > 1$, then $|1/z| < 1$, so we can expand

$\frac{1}{1-z} = -\frac{1}{z} \cdot \frac{1}{1-\frac{1}{z}}= -\frac{1}{z} (1 + (\frac{1}{z} + \frac{1}{z}^2 + \cdots) = \frac{1}{z} - (\frac{1}{z})^2 - (\frac{1}{z})^3 - \cdots$

for $|z| > 1$. Since this has infinitely many terms in the Laurent series, $z=1$ is an essential singularity.

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If the book really says that, it's quite sad. Consider writing them to tell them about this egregious error. –  Gerry Myerson Aug 31 '11 at 7:09
One hopes that was not one of the author's better days. –  Michael Hardy Aug 31 '11 at 13:28
Or the OP has momentarily confused "nonremovable" with "essential". They are synonyms in some other contexts. –  Henning Makholm Aug 31 '11 at 14:15

1 Answer 1

I don't think you are missing anything.

$f(z)=\frac{1}{1-z}$ has a pole of order one at $z=1$. As an extra reassurence, perhaps look at the definition of a pole as is on Wikipedia. In our case $g(z)=-1$.

Added: This is to answer the updated question:

When we talk about the Laurent series centered at $a$, we are talking about something of the form $\sum_{k} c_k (z-a)^k$. The series you wrote above is centered at $0$, not $1$, so it is not the Laurent expansion at $z=1$. However, there is another question we need to answer: Why doesn't this mean our function has an essential singular at $z=0$? Here the problem is that there exists an open neighbourhood of $1$ where that series does not apply. Precisely, it does not converge for any $|z|<1$.

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