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

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.

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

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

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. $$z=1$$: $$$$f(z)\left( z-i\right) =\frac{1}{z^{2}}=\frac{A}{z^{2}}\left( z-i\right) +\frac{B}{z}\left( z-i\right) +C,$$$$

$$$$z=1\implies 1=i\left( 1-i\right) +B\left( 1-i\right) -1\implies B=1.$$$$

Then

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

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

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

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

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

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 $$$$g_{1}(w)=\frac{1}{w}\sum_{n\geq 0}\frac{\left( -1\right) ^{n}}{w^{n}} =\sum_{n\geq 0}\frac{\left( -1\right) ^{n}}{w^{n+1}}, \text{ for } \color{blue}{1<\left\vert w\right\vert} . \tag{5}$$$$
2. As for the second term, for $$\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 $$\color{blue}{1<}\left\vert z-1\right\vert \color{green}{<\sqrt{2}}$$:

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

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.