How can the Laurent series for 1/(1+z^2) be found around z=i?

[vertigo74]

Find the Laurent series that converges for 0 < | z - i| < R of

\frac {1}{1 + z^2}

I have been given the hint to break it up as

\frac {1}{1 + z^2} = (\frac {1}{z - i})(\frac {1}{z + i}) and then expand \frac {1}{z + i} . I am kind of confused about this, because the series is centered at $i$. I'm not exactly sure how to do it because the center isn't 0.

The solution is -\sum_{n = 0}^{\infty}(\frac {i}{2})^{2n + 1}(z - i)^{n - 1}

 

[Avodyne]

You could let z = w+i, and expand in powers of w. Then, at the end, set w = z-i.

 

[HallsofIvy]

You learned a long time ago that the general Taylor's series (as opposed to Maclaurin series) is about "x= a" rather than "x= 0". You want to expand \frac{1}{z+i}[/itex] around z= i: in powers of (z- i). You could do that by taking derivatives and doing an actual Taylor's series expansion, evaluating the derivatives at z= i rather than at z= 0.