My task is to find the Laurent series expansion of $$f(z)=\frac{1}{2z^2-z-1},$$ which converges for $\displaystyle \frac12 <|z|<1$.

I proceeded by doing the following: $$\frac{1}{2z^2-z-1} = \frac{1}{(2z+1)(z-1)} = \frac13 \left( \frac{1}{z-1}-\frac{2}{2z+1} \right).$$

For the first term: $$\frac13 \frac{1}{z-1} = \frac{1}{3z} \frac{1}{1-\frac1z}=\frac{1}{3z}\sum_{n=0}^\infty \frac{1}{z^n},$$ for $\displaystyle \left| \frac1z \right|<1 \Rightarrow |z|>1.$ Hence, for $|z|<1$ (using geometric series properties): $$\frac13 \frac{1}{z-1} =-\frac{1}{3z}\sum_{n=1}^\infty z^n=-\frac13 \sum_{n=1}^\infty z^{n-1}.$$ For the second term: $$-\frac13\frac{2}{2z+1}=-\frac23\frac{1}{1+2z}=-\frac23\sum_{n=0}^\infty\left( -2z \right)^n,$$ for $\displaystyle |2z|<1 \Rightarrow |z|<\frac12$. Hence, for $\displaystyle |z|>\frac12$: $$-\frac13\frac{2}{2z+1}=\frac23 \sum_{n=1}^\infty\frac{1}{(-2z)^n}=\frac13 \sum_{n=1}^\infty(-1)^{-n}2^{1-n}z^{-n}.$$ Hence, $$\frac{1}{2z^2-z-1}=\frac13 \sum_{n=1}^\infty(-1)^{-n}2^{1-n}z^{-n}-\frac13 \sum_{n=1}^\infty z^{n-1}.$$

I think this is the correct approach. Some further guidance/clarification would be great!


1 Answer 1


Note: This is just an affirmation of OPs calculation.

The function

\begin{align*} f(z)&=\frac{1}{3}\frac{1}{1-z}-\frac{1}{3}\frac{1}{z+\frac{1}{2}}\\ \end{align*} has two simple poles at $z=1$ and $z=-\frac{1}{2}$.

Since we want to find a Laurent expansion with center $0$, we look at these poles and see they determine three regions.

\begin{align*} |z|<\frac{1}{2},\qquad\quad \frac{1}{2}<|z|<1,\qquad\quad 1<|z| \end{align*}

  • The first region $ |z|<\frac{1}{2}$ is a disc with center $0$, radius $\frac{1}{2}$ and the pole $\frac{1}{2}$ at the boundary of the disc. In the interior of this disc all two fractions with poles $\frac{1}{2}$ and $1$ admit a representation as power series at $z=0$.

  • The second region $\frac{1}{2}<|z|<1$ is the annulus with center $0$, inner radius $\frac{1}{2}$ and outer radius $1$. Here we have a representation of the fraction with pole $\frac{1}{2}$ as principal part of a Laurent series at $z=0$, while the fraction with pole at $1$ admits a representation as power series.

  • The third region $|z|>1$ containing all points outside the disc with center $0$ and radius $1$ admits for all fractions a representation as principal part of a Laurent series at $z=0$.

A power series expansion of $\frac{1}{z+a}$ at $z=0$ is \begin{align*} \frac{1}{z+a}&=\frac{1}{a}\cdot\frac{1}{1+\frac{z}{a}}\\ &=\sum_{n=0}^{\infty}\frac{(-1)^n}{a^{n+1}}z^n \end{align*} The principal part of $\frac{1}{z+a}$ at $z=0$ is \begin{align*} \frac{1}{z+a}&=\frac{1}{z}\cdot\frac{1}{1+\frac{a}{z}}=\frac{1}{z}\sum_{n=0}^{\infty}\frac{(-a)^n}{z^n}\\ &=\sum_{n=1}^{\infty}\frac{(-a)^{n-1}}{z^n} \end{align*}

OPs calculation corresponds to the representation in the second region and is correct. It is sufficient to calculate the power series part for $\frac{1}{1-z}$ and the principal part of the Laurent series for $\frac{2}{2z+1}$.


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