# Question regarding singularity of a complex function

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?

• 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 $f(5i)=-1/14$
– mike
Commented Jul 15, 2014 at 21:30
• @mike I have edited the question, I think you were right, there was a mistake, now seems to be correct, thanks! Commented Jul 15, 2014 at 22:04
• 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.] Commented Jul 15, 2014 at 22:13

## 1 Answer

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 $g(w) (f(z))$.

• 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 Commented Jul 15, 2014 at 21:08