Consider the function $$f(z) = {1 \over (z-i)(z+i)}$$

with a Laurent series expansion at $z_0=i$ on a domain $\;\Omega=\left\{z\in \mathbb{C}:2\lt\left|z-i\right|\right\}$ $$\begin{eqnarray}f(z)={1 \over (z-i)(z+i)}={1 \over (z-i)}{1 \over (z+i)}={1 \over (z-i)}{1\over(z-i)+2i} \\={1 \over (z-i)}{1 \over (z-i)}{1\over1-\left(-{2i\over z-i}\right)}\\={1 \over (z-i)^2}\sum_{n=0}^\infty\left(-{2i\over z-i}\right)^n\\=\sum_{n=0}^\infty (-2i)^n\left({1\over z-i}\right)^{n+2}\\=\sum_{n=-\infty}^{-2}\left(\frac i2\right)^{n+2}(z-i)^n\end{eqnarray}$$

Just to make sure I'm getting everything correct. Obviously we have a singularity at $z=i$, however cannot determine its type, as $i\notin \Omega$. Is it true that we can't determine the type of this singularity on the restricted domain as it doesn't lie in this domain?

  • $\begingroup$ Please double-check your expansion formula. It seems not correct to me. If we set $z=5i$, then your function $f(5i)=-1/24$, but your expansion $f(5i)=-1/14$ $\endgroup$
    – mike
    Commented Jul 15, 2014 at 21:30
  • $\begingroup$ @mike I have edited the question, I think you were right, there was a mistake, now seems to be correct, thanks! $\endgroup$ Commented Jul 15, 2014 at 22:04
  • $\begingroup$ Right, the Laurent series for $\lvert z-i\rvert > 2$ doesn't (directly) tell you what sort of singularity there is at $i$. [Of course you can obtain an expression for the function that does tell you that from this Laurent series.] $\endgroup$ Commented Jul 15, 2014 at 22:13

1 Answer 1


Set $z=i+w$, then

$$f(z) = g(w)={1 \over w(w+2i)}=\frac{1}{2iw (1+(w/2i))}=\frac{1}{2iw}\left(1-(w/2i)+(w/2i)^2-(w/2i)^3+\cdots\right)$$

So $w=0$($z=i$) is a pole of $g(w) (f(z))$.

  • $\begingroup$ Hey mike. The point is that $i$ doesn't belong to the restricted domain, so we cannot determine the type of this singularity with the current Laurent expansion, as it corresponds to the function at this domain $\endgroup$ Commented Jul 15, 2014 at 21:08

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