Just checking, Complex Analysis, Series? 10 points for informative answer!

Hello Experts,

I have 1/z and I need to expand it to series at z=1. I know that the region is: |z-2|+|z+2|<6

Now I know that d^n/dz [1/z] at z=1 is (-1)^n n=0 to infinity.
The function is analytic therefore I can expand it to Taylor series at z=1:

Sigma from n=0 to infinity of (-1)^n/n! * (z-1)

Please correct me if I am mistaken.

For the Laurent Series of 1/z at z = 1:

Using the geometric series,
1/z = 1/[1 - (-(z - 1))]
......= 1 - (z - 1) + (z - 1)^2 - (z - 1)^3 + ...
......= Σ(k = 0 to ∞) (-1)^k (z - 1)^k.

Note:
(d^n/dz^n) (1/z) = (-1)^n * n! / z^(n+1); don't forget the factorial! (pun intended...)
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I hope this helps!