# Lauarent Series of 1/(1+z^2)

1. Nov 6, 2007

### 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 -$$\sum_{n = 0}^{\infty}(\frac {i}{2})^{2n + 1}(z - i)^{n - 1}$$

2. Nov 6, 2007

### Avodyne

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

3. Nov 6, 2007

### HallsofIvy

Staff Emeritus
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 [tex]\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.

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