I'm really confused by the Laurent series, and need a little help with this one:

$ f(z) = \frac z{(z-6i)^9} $ around $z = 6i$

The partial fraction method doesn't seems adapted for this problem. I wrote $$ f(z) = \sum_{n=-\infty}^{\infty} a_{n} (z-6i)^{n} $$ With $$a_{n} = \frac1 {2\pi i} \oint_C \frac {f(z)}{(z-6i)^{n+1}} dz$$

$$ f(z) = \sum_{n=-\infty}^{\infty} \frac1 {2\pi i} \oint_C \frac z{(z-6i)^{n+10}} dz . (z-6i)^{n} $$ I don't know what to do next. I'm also supposed to tell in which region the expansion is valid. As there's only one pole in $z=6i$ I was thinking that the expansion would be valid in the whole complex plane?



1 Answer 1


Hint: You can write $z$ as $z-6i + 6i$. Then $$\frac{z}{(z-6i)^9} = \frac{z-6i}{(z-6i)^9} +\frac{6i}{(z-6i)^9} $$ $$= \frac{6i}{(z-6i)^9}+\frac{1}{(z-6i)^8} $$ This looks valid for all $z\neq 6i$ to me. The series is finite.

  • $\begingroup$ Thanks for the quick answer. But in this case, I don't understand the utility of the Laurent series, because we still don't have any value for $f(6i)$ $\endgroup$
    – Sihtam
    Commented Jan 11, 2015 at 19:14
  • $\begingroup$ The function has a pole at $6i$. There is no value there because $6i$ is not in the domain of the function. It's often easiest to obtain the Laurent series as a sort of trick or manipulation of another series or expression, instead of direct computation. $\endgroup$
    – MPW
    Commented Jan 12, 2015 at 12:20

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