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Let $f(z)= \frac {1 } {z^3+z^2 } $ and consider its series expansions around the point $z=1 $.

Clearly there will be three such, on 1) $|z-1 |< 1 $, 2) $1<|z-1 |<2 $, 3) $|z-1 |>2$.

Now my question: how can I tell which of these will be Taylor series or which will be Laurent series? Obviously one way would be to partial fraction the expression and then express the fractions as series using the fact that $\frac {1 } {1-z } =\sum _{n=0 } ^{\infty } z^n, |z |<1$, and then see if the series contains negative exponents. But is there an easier way to determine this without doing this work?

Thanks in advance!

Loreno Heer
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Alexander
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1 Answers1

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If the Laurent series of $f$ in the annulus $r < \lvert z-a\rvert < R$ is actually a Taylor series (i.e., it contains no negative powers of $z-a$), then the series converges (at least) on the disk $\lvert z-a\rvert < R$, and therefore $f$ can have no singularities in that disk.

Thus in the given example, the Laurent series will be a Taylor series only for the disk $\lvert z-1\rvert < 1$, since $f$ has poles in $0$ and $-1$.

Daniel Fischer
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  • (More trivially, if a Laurent series converges on a disk, it must actually be a Taylor series...) – Micah Jul 04 '14 at 20:59
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    Perhaps important to add: By analytic continuation, $f$ coincides with any of its series expansions wherever the series converges. So if the Laurent series in the disk $1<|z-1|<2$ were actually a Taylor series, it sum would coincide with $f$ in the entire disk $|z-1|<2$, which is impossible because of the pole at $z=0$. – Harald Hanche-Olsen Jul 04 '14 at 21:03
  • Ok thanks, as I though, but there is an answer to this exercise wich say that for $|z-1 |>2 $ there is "a convergent power series in $\frac {1 } {z-1 } $". What is meant by that? – Alexander Jul 04 '14 at 21:09
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    I can't know for sure, what is meant, @Alexander, but probably the meaning is that the Laurent series in $\lvert z-1\rvert > 2$ contains no positive powers of $(z-1)$. [The function has a removable singularity - a zero - in $\infty$, hence the Laurent series in the annulus $\lvert z-1\rvert > 2$ consists only of its principal part, which can be called a "power series" in $(z-1)^{-1}$.] – Daniel Fischer Jul 04 '14 at 21:18