can someone help with the laurant expansion of
$$\frac{1+z}{z}$$ at 0? i get
$$2+\sum_{k=1}^{\infty}(-1)^k (z-1)^k$$ but i need the expansion for all z

Question

can someone help with the laurant expansion of 
$$\frac{1+z}{z}$$ at 0?  i get 
$$2+\sum_{k=1}^{\infty}(-1)^k (z-1)^k$$ but i need the expansion for all z

jamesj · Answer

Do you mean the Laurant expansion at z = 1 ?

anonymous · Answer

maybe i am just being daft. maybe it is only 
$$\frac{1}{z}+1$$ and i am thinking way too hard?

jamesj · Answer

At z = 0, the expansion is just 1/z + 1, yes.

anonymous · Answer

ok i get the dope slap award.  i was trying to do it as if it was a geometric series, but now i am embarrassed.

jamesj · Answer

live and learn.  Or calculate and learn.

anonymous · Answer

well here is one that i am actually going to have to think about.  expand 
$$\frac{z}{z^2+1}$$ about i on 
$$|z-i|<2$$ any suggestions?

anonymous · Answer

partial fractions?

jamesj · Answer

$$ \frac{z}{z^2 + 1} = \frac{1}{z-i} \frac{z}{z+i} $$
So if you have an expansion of the z/(z+i) term, you you just shift all the terms down by one power of (z-i).

Now, for the z/(z+i) itself, you can start differentiating and evaluating at i, or write it as
$$ \frac{z}{z+i} = 1 - \frac{i}{z+i} $$
and start differentiating and evaluating that, which is very manageable.

anonymous · Answer

ok thank you so much.