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?

  • $\begingroup$ 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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.