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.

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-iright) }=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-iright) $ and use the root $z=i$ of the denominator $q(z)= z^{2}left( z-iright)$ and evaluate the limit begin{equation*} C=lim_{zrightarrow i}f(z)left( z-iright) =lim_{zrightarrow 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_{zrightarrow 0}f(z)z^{2}=lim_{zrightarrow 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-iright) $ or $z^{2}$ , and pick a meaningful $z$, e.g. $z=1$: begin{equation} f(z)left( z-iright) =frac{1}{z^{2}}=frac{A}{z^{2}}left( z-iright) +frac{B}{z}left( z-iright) +C, end{equation}

begin{equation} z=1implies 1=ileft( 1-iright) +Bleft( 1-iright) -1implies B=1. end{equation}

Then

begin{equation} f(z)equiv frac{1}{z^{2}left( z-iright) }=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 $1<leftvert z-1rightvert <sqrt{2}$ becomes the new annulus $1<leftvert wrightvert <sqrt{2}$, centered at $w=0$, and $frac{1}{z^{2}left( z-iright) }$ becomes

begin{equation} frac{1}{z^{2}left( z-iright) }=frac{1}{left( w+1right) ^{2}left[ w+left( 1-iright) 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+1right) ^{2}}-frac{1}{w+left( 1-iright) }. tag{4} end{equation}

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

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

D. From $(5)-(7)$ it follows that for $color{blue}{1<}leftvert wrightvert color{green}{<sqrt{2}}, $

begin{align} g(w) &=g_{1}(w)+g_{2}(w)+g_{3}(w) \ &=sum_{ngeq 0}left( -1right) ^{n}left[ frac{1-in}{w^{n+1}}+frac{w^{n}}{left( 1-iright) ^{n+1}}right] text{ for }color{blue}{1<}leftvert wrightvert 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<}leftvert z-1rightvert color{green}{<sqrt{2}} $:

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

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 thatbegin{align}frac1{z^2}&=-left(frac1zright)'\&=-left(sum_{n=-infty}^{-1}(-1)^n(z-1)^nright)'\&=-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 $|z-1|<sqrt2$, thenbegin{align}-frac1{z-i}&=frac1{i-z}\&=frac1{-1+i-(z-1)}\&=frac1{-1+i}sum_{n=0}^inftyfrac{(z-1)^n}{(-1+i)^n}\&=sum_{n=0}^inftyfrac{(z-i)^n}{(-1+i)^{n+1}}.end{align}So, all that remains to be done is to put these three series together.