Laurent Series for $\frac{1}{1 – z}$ and $\frac{1}{(1 – z)^2}$ around $|z| > 1$.

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I am supposed to find the Laurent series for $f(z) = \frac{1}{1 – z}$ and $ f(z) = \frac{1}{(1 – z)^2}$ around $|z| > 1$. I think I can find the first one, but the second one to me seems a little bit confusing.

For the first, one since we want a Laurent series that converges for $|z| > 1$, i.e. $|1/z| < 1$, then we can write the given function as
\begin{equation*}
f(z) = -\dfrac{1}{z}\dfrac{1}{1 – \frac{1}{z}} = -\dfrac{1}{z} \sum_{n = 0}^{\infty} \left(\dfrac{1}{z}\right)^n = -\sum_{n = 0}^{\infty} \left(\dfrac{1}{z}\right)^{n + 1}
\end{equation*}
which is the desired series that will converge for $|z| > 1$.

Now for the second one, I know that I can write the function as
\begin{equation*}
f(z) = \dfrac{1}{z^2}\dfrac{1}{\left(1 – \frac{1}{z}\right)^2}
\end{equation*}
But I am not sure how I would be able to expand $\frac{1}{\left(1 – \frac{1}{z}\right)^2}$ as a series. I am not sure if it would follow from the first expansion, or not.

Some help would be appreciated.

Best Answer

Besides the nice answer from @eyeballfrog you can also directly obtain the Laurent series by applying a binomial series expansion admissible for $\left|\frac{1}{z}\right|<1$.

We obtain \begin{align*} \frac{1}{z^2}\frac{1}{\left(1 - \frac{1}{z}\right)^2} &=\frac{1}{z^2}\sum_{n=0}^\infty\binom{-2}{n}\left(-\frac{1}{z}\right)^n\\ &=\frac{1}{z^2}\sum_{n=0}^\infty(n+1)\frac{1}{z^n}\tag{1}\\ &\,\,\color{blue}{=\sum_{n=2}^\infty(n-1)\frac{1}{z^{n}}}\tag{2} \end{align*}

Comment:

  • In (1) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

  • In (2) we shift the index by $2$ to get an expansion in terms of $\frac{1}{z^n}$.

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