Is this sum correct?
$$
\frac{z^2-2z+3}{z-2}=((z-1)^2+2)\sum_{j=0}^\infty \frac{1}{(z-1)^{j+1}}
$$
$$
=(z-1)+1+\sum_{j=0}^\infty \frac{3}{(z-1)^{j+1}}
$$
I don't understand how you go from the first line to the second line.
Laurent series sum for function $ \frac{z^2-2z+3}{z-2} $
0
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complex-analysis
laurent-series
1 Answers
1
Hints:
first equality follows from $z^2-2z+3=(z-1)^2+2\,$
second equality follows from
$$ \begin{align} \big((z-1)^2+2\big)\sum_{j=0}^\infty \frac{1}{(z-1)^{j+1}} &= \sum_{j=0}^\infty \frac{1}{(z-1)^{j-1}}+\sum_{j=0}^\infty \frac{2}{(z-1)^{j+1}} \\ & = (z-1)+1+\sum_{j=2}^\infty \frac{1}{(z-1)^{j-1}}+\sum_{j=0}^\infty \frac{2}{(z-1)^{j+1}} \end{align} $$