# Laurent Series 1/(z-3i)

Apologies if this is a simple question. I've just read a chapter on Laurent series which seems to indicate that in practice we don't calculate the coefficients of a Laurent series using the integral, instead we often manipulate Taylor series, adjust a function so a geometric expansion may be used etc.

For the function:

\begin{align} f(z) = \frac{1}{z-3i} \end{align}

How do we obtain the Laurent series at about 3i, i.e. how do we obtain:

\begin{align*} \frac{1}{z-3i} + 0+ 0(z-3i) + 0(z-3i)^{2} + \cdots \quad z \in \mathbb{C}-\{3i\} \end{align*}

Many thanks,

John

• You already have the Laurent series for $f(z)$, it's just itself, since you can write $f$ as $f(z) = \sum_{n=-\infty}^{+\infty} a_n (z-3i)^n$, with $a_n = 0$ for $n \neq -1$ and $a_{-1} = 1$ – Klaramun Nov 12 '14 at 10:34
• That makes complete sense. Many thanks. – John Smith Nov 12 '14 at 10:49
• Actually, it doesn't make sense. I've seen the definition of an extended power series, but the way it's applied seems strange, as if we don't actually calculate anything? I have the same lack of understanding with $\frac{1}{(z+i)^{2}} = \frac{1}{(z+i)^{2}} + \frac{0}{(z+i)} + \cdots$. How do we know the coefficients are zero of the other terms? – John Smith Nov 12 '14 at 11:00
• You don't need to compute the coefficients because it is already a Laurent series. You should compute the coefficients if for example you function would be $f(z) = \frac{1}{z^2 - 1}$, because it is not of the form above. – Klaramun Nov 12 '14 at 11:07
• Read about uniqueness of the Laurent representation. – Pedro Tamaroff Nov 12 '14 at 11:21