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Is the Laurent expansion in $1<\vert z \vert <\infty$ equivalent to the Laurent expansion at $\infty$? If so, $$f(1/z)=z/(1+1/z^2)=z^3/(1-(-z^2))=z^3-z^5+z^7-z^9+\cdots$$

However, wolframalpha says the expansion at infinity is $1/z^3-1/x^5-1/z^7+\cdots$

When I am asked to expand a Laurent series outside of some disc or annulus, should I just find the expansion at infinity?

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  • $\begingroup$ Why do you use $f(1/z)$) $\endgroup$ – Zelos Malum Oct 26 '15 at 5:37
  • $\begingroup$ Because I thought the expansion of $f(z)$ at infinity is the expansion of $f(1/z)$ at $0$. $\endgroup$ – cap Oct 26 '15 at 5:37
  • $\begingroup$ While I am fairly new to complex analysis I'd say you probably shouldn't do that as it is not at infnity, but IF we go by it and I assume you wrote wolfram wrong as it has $x$ in it, they are identical, because your $z$ is near 0 while htiers is "near" infinity, which means you must invert your $z$ such that they get negative exponents. $\endgroup$ – Zelos Malum Oct 26 '15 at 5:39
  • $\begingroup$ That's what I am unsure of, what is my expansion centered at? $\endgroup$ – cap Oct 26 '15 at 5:40
  • $\begingroup$ From what you said i'd say $0$ while theirs is at $\infty$ if we go by that, which of course are each others inverse on the riemannsphere and all so just inverse the exponents. $\endgroup$ – Zelos Malum Oct 26 '15 at 5:41
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You want an expansion of $f$ for $|z|>1$.

An easy way is to write $f(z) = {1 \over z^3} {1 \over 1+ {1 \over z^2}}$.

Since $|z|>1$, we have ${1 \over |z|^2 } < 1$ and so ${1 \over 1+ {1 \over z^2}} = 1 -{ 1\over z^2} + {1 \over z^4} - \cdots $.

Hence $f(z) = \sum_{n=0}^\infty (-1)^n {1 \over z^{n+3}}$.

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