How do I write the Laurent series for $\frac{1}{z^2(z-i)}$ for $1<|z-1|<\sqrt2$?

Question

How do I write the Laurent series for $\frac{1}{z^2(z-i)}$ for $1<|z-1|<\sqrt2$?

I know that I need to rewrite it somehow to fit the geometric series form of $\frac{1}{1-r}$, but I'm stuck on getting here. I'm also aware that $z_0=1$, so then my "r" should have $z-1$ in it correct? After that I think I can do the rest from there.

Answer 1

I know that I need to rewrite it somehow to fit the geometric series form of $\frac{1}{1-r}$, but I'm stuck on getting here.

If you know already that you should use partial fractions, then you can jump to Equation $(2)$ and from there on. In B. and C. the geometric series you need are indicated and motivated, I hope.

I'm also aware that $z_0=1$, so then my "r" should have $z-1$ in it correct?

Yes.

A. When the given function is of the form $f(z)=\frac{p(z)}{q(z)}$, with $p(z)$ and $q(z)$ being polynomials in $z$, the first step is to expand it into partial fractions. Due to the form of $f(z)$ this means that

\begin{equation} f(z)\equiv \frac{1}{z^{2}\left( z-i\right) }=\frac{A}{z^{2}}+\frac{B}{z}+\frac{C}{z-i}. \tag{1} \end{equation}

To find the coefficients we can use the Heaviside cover-up method.

• To determine $C$, multiply by $\left( z-i\right) $ and use the root $z=i$ of the denominator $q(z)= z^{2}\left( z-i\right)$ and evaluate the limit \begin{equation*} C=\lim_{z\rightarrow i}f(z)\left( z-i\right) =\lim_{z\rightarrow i}\frac{1}{z^{2}}=\frac{1}{i^{2}}=-1. \end{equation*}

• To find $A$, multiply by $z^{2}$ and use the root $z=0$ of $q(z)$: \begin{equation*} A=\lim_{z\rightarrow 0}f(z)z^{2}=\lim_{z\rightarrow 0}\frac{1}{z-i}=i. \end{equation*}

• To determine $B$, substitute $C$ and $A$ in one of the equations resulting from $(1)$ after being multiplied by $\left( z-i\right) $ or $z^{2}$ , and pick a meaningful $z$, e.g. $z=1$: \begin{equation} f(z)\left( z-i\right) =\frac{1}{z^{2}}=\frac{A}{z^{2}}\left( z-i\right) +\frac{B}{z}\left( z-i\right) +C, \end{equation}

\begin{equation} z=1\implies 1=i\left( 1-i\right) +B\left( 1-i\right) -1\implies B=1. \end{equation}

Then

\begin{equation} f(z)\equiv \frac{1}{z^{2}\left( z-i\right) }=\frac{i}{z^{2}}+\frac{1}{z}- \frac{1}{z-i}. \tag{2} \end{equation}

B. To make some algebraic manipulations easier we now use the substitution $w=z-1$. Then the annulus $1<\left\vert z-1\right\vert <\sqrt{2}$ becomes the new annulus $1<\left\vert w\right\vert <\sqrt{2}$, centered at $w=0$, and $\frac{1}{z^{2}\left( z-i\right) }$ becomes

\begin{equation} \frac{1}{z^{2}\left( z-i\right) }=\frac{1}{\left( w+1\right) ^{2}\left[ w+\left( 1-i\right) \right] }\equiv g(w). \tag{3} \end{equation}

By $(2)$ the function $g(w)$ can be expanded as

\begin{equation} g(w)=\frac{1}{w+1}+\frac{i}{\left( w+1\right) ^{2}}-\frac{1}{w+\left( 1-i\right) }. \tag{4} \end{equation}

C. Each term can be expanded into a particular geometric series as follows:

1. For $\color{blue}{1<\left\vert w\right\vert }$ and using the sum of the complex geometric series $\displaystyle\sum_{n\geq 0}\dfrac{\left( -1\right) ^{n}}{w^{n}}=\frac{1}{1-\left(-1/w\right) }$, the first term can be written as \begin{equation*} g_{1}(w)\equiv \frac{1}{w+1}=\frac{1}{w}\frac{1}{1+1/w}=\frac{1}{w}\frac{1}{ 1-\left( -1/w\right) } \end{equation*} and expanded into \begin{equation} g_{1}(w)=\frac{1}{w}\sum_{n\geq 0}\frac{\left( -1\right) ^{n}}{w^{n}} =\sum_{n\geq 0}\frac{\left( -1\right) ^{n}}{w^{n+1}}, \text{ for } \color{blue}{1<\left\vert w\right\vert} . \tag{5} \end{equation}
2. As for the second term, for $\color{blue}{\left\vert w\right\vert >1}$ as well, since $$\frac{1}{\left( w+1\right) ^{2}}=-\frac{d}{dw}\left( \frac{1}{w+1}\right) =-\frac{d}{dw}g_{1}(w),$$ it can be expanded into \begin{align} g_{2}(w) &\equiv \frac{i}{\left( w+1\right) ^{2}}=-i\frac{d}{dw}g_{1}(w)=-i\frac{d}{dw}\sum_{n\geq 0}\frac{\left( -1\right) ^{n}}{w^{n+1}} \\ &=i\sum_{n\geq 0}\left( -1\right) ^{n}\frac{n+1}{w^{n+2}}=-i\sum_{n\geq 1}\left( -1\right) ^{n}\frac{n}{w^{n+1}},\text{ for }\color{blue}{1<\left\vert w\right\vert }. \tag{6} \end{align}
3. As for the third term, if $\left\vert - \dfrac{w}{1-i}\right\vert =\color{green}{\dfrac{\left\vert w\right\vert }{\sqrt{2}}<1}$, we have that \begin{align} g_{3}(w) &\equiv \frac{1}{w+\left( 1-i\right) }=\frac{1}{1-i}\frac{1}{1-\left( -\frac{w}{1-i}\right) } \\ &=\frac{1}{1-i}\sum_{n\geq 0}\frac{\left( -1\right) ^{n}w^{n}}{\left( 1-i\right) ^{n}}=\sum_{n\geq 0}\frac{\left( -1\right) ^{n}w^{n}}{\left(1-i\right) ^{n+1}}\text{ for }\color{green}{\left\vert w\right\vert <\sqrt{2}}. \tag{7} \end{align}

D. From $(5)-(7)$ it follows that for $\color{blue}{1<}\left\vert w\right\vert \color{green}{<\sqrt{2}}, $

\begin{align} g(w) &=g_{1}(w)+g_{2}(w)+g_{3}(w) \\ &=\sum_{n\geq 0}\left( -1\right) ^{n}\left[ \frac{1-in}{w^{n+1}}+\frac{w^{n}}{\left( 1-i\right) ^{n+1}}\right] \text{ for }\color{blue}{1<}\left\vert w\right\vert \color{green}{<\sqrt{2}}. \tag{8} \end{align}

In terms of the given function $f(z)$, we thus have the following expansion for $\color{blue}{1<}\left\vert z-1\right\vert \color{green}{<\sqrt{2}} $:

\begin{equation} f(z)=\sum_{n\geq 0}\left( -1\right) ^{n}\left[ \frac{1-in}{\left( z-1\right) ^{n+1}}+\frac{\left( z-1\right) ^{n}}{\left( 1-i\right) ^{n+1}}\right] \text{ for }\color{blue}{1<}\left\vert z-1\right\vert \color{green}{<\sqrt{2}} . \tag{9} \end{equation}

Answer 2

First of all, you should do use the fact that$$\frac1{z^2(z-i)}=\frac i{z^2}-\frac1{z-i}+\frac1z.$$So, if $|z-1|>1$, \begin{align}\frac1z&=\frac1{1+(z-1)}\\&=-\sum_{n=-\infty}^{-1}(-1)^n(z-1)^n.\end{align}It follows from this that\begin{align}\frac1{z^2}&=-\left(\frac1z\right)'\\&=-\left(\sum_{n=-\infty}^{-1}(-1)^n(z-1)^n\right)'\\&=-\sum_{n=-\infty}^{-1}n(-1)^n(z-1)^{n-1}\\&=\sum_{n=-\infty}^{-2}(n+1)(-1)^n(z-1)^n.\end{align}On the other hand, if $|z-1|<\sqrt2$, then\begin{align}-\frac1{z-i}&=\frac1{i-z}\\&=\frac1{-1+i-(z-1)}\\&=\frac1{-1+i}\sum_{n=0}^\infty\frac{(z-1)^n}{(-1+i)^n}\\&=\sum_{n=0}^\infty\frac{(z-i)^n}{(-1+i)^{n+1}}.\end{align}So, all that remains to be done is to put these three series together.