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How can we compute the Laurent expansion of

$$f(z)=\frac{1}{z(z-i)^2}$$

about $0$ in $\Omega=B_1(0)\setminus\{0 \}$?

Thoughts:

By partial fractions we have that

$$f(z)=\frac{1}{z-i}-\frac{i}{(z-i)^2}-\frac{1}{z}$$

but I'm unsure is to which expansions will be valid in $\Omega$. Any help would be very appreciated.

Best, MM.

1 Answers 1

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here's a trick: $$ \frac{1}{z-i}=\frac{i}{1-z/i}=i\sum_{n=0}^{\infty}(z/i)^n $$ also, $$ \frac{d}{dz}\frac{1}{z-i}=\frac{-1}{(z-i)^2} $$ so that $$ \frac{-i}{(z-i)^2}=i\frac{d}{dz}\frac{1}{z-i}=i\frac{d}{dz}i\sum_{n=0}^{\infty}(z/i)^n=-\sum_{n=1}^{\infty} nz^{n-1}(-i)^n $$ you can piece it together from there

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    Many thanks! Exactly what I was looking for.2012-01-14