Laurent series
Expand the function f (z) into the Laurent series at point z0 for the given annulus.
f (z) = 1 / ((z-2) * (z-3))
a) z0 = 2, 0 <| z-2 | <1
b) z0 = 2, | z-2 |> 1
c) z0 = infinity, | z |> 3
Your answer:
f (z) = 1 / ((z-2) * (z-3))
a) z0 = 2, 0 <| z-2 | <1
b) z0 = 2, | z-2 |> 1
c) z0 = infinity, | z |> 3
Your answer:
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