9)Find Laurent series expansion of the function 1/ z²+z-6 in the regions
1<|z-1 <4 and |z-1|> 4.
From partial fraction
"\\frac{1}{(z^2+z-6)}=\u2212\\frac{1}{z\u22121}+\\frac{1}{z\u22122}\\\\"
so on either annulus, it is enough to find the Laurent series expansions of the functions "\\frac{1}{z\u22121}" and "\\frac{1}{z\u22122}" and then combine them term by term.
If "|z|>2" then note that
"\\frac{1}{z\u22122}=\\frac{1}{z} \\frac{1}{1\u22122z^{\u22121}}"
and the fact that |z|>2
|z|>2 implies that |2z
−1
|<1
|2z−1|<1. If you recall the geometric series formula,
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