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I'm trying to find the Laurent series expansion of $f(z)=\frac{1}{z}$ around $z=1+i$ (i.e. the series will be in terms of $z-1-i$), but cannot seem to do it. Can anyone help?

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Did you mean "around $\,z=1-i\,$"? –  DonAntonio Feb 14 '13 at 3:21
    
Sorry, I mean around $1+i$ so the resulting series is in terms of $z-i-1$ –  user62245 Feb 14 '13 at 3:24

1 Answer 1

You have $$\frac1z=\frac{1}{(1+i)+\bigl(z-(1+i)\bigr)}=\frac1{1+i}\,\cdot\,\frac{1}{1+\frac{(z-(1+i))}{1+i}}\,,$$

which can be expanded as a geometric series, provided that $$\biggl|\frac{z-(1+i)}{1+i}\biggr|<1\,,$$

hence

$$\frac1z=\sum_{n=0}^\infty(-1)^n\frac1{(1+i)^{n+1}}\,\bigl(z-(1+i)\bigr)^n\,.$$

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