# Laurent series around z=-i for $\frac{1}{z(z-1)}$ and $1<|z+i|<\sqrt2$

I'm given the function :

$$f(z)=\frac{1}{z(z-1)}$$

I'm interesting in finding the Laurent series around $z=-i$ for the one finite circular ring corresponding to this function given its singularities ( I hope I'm using the right words here).

I have some ideas but having seen very few problems, I'm not sure about them. So this question is mostly here to check if the way I do things is correct.

I think the area we are interested in is the following : $1<|z+i|<\sqrt2$

$f(z)$ can be written as $f(z)=\frac{1}{z-1}-\frac{1}{z}$

We can write the first term as $$\frac{1}{z-1}=\frac{-1}{(1+i)(1-\frac{z+i}{1+i})}$$ $$\frac{-1}{(1+i)}\frac{1}{(1-\frac{z+i}{1+i})}=\frac{-1}{(1+i)}\sum_{n=0}^\infty\left(\frac{z+i}{1+i}\right)^n$$

I tried to turn this in the form of the geometric series.

This is valid for $$\frac{|z+i|}{|1+i|}<1=>|z+i|<\sqrt2$$

For the other part of the inequality I will try to do the same for the second term of $f(z)$ :

$$\frac{1}{z}=\frac{1}{(z+i)(1-\frac{i}{z+i})}=\frac{1}{z+i}\sum_{n=0}^\infty\left(\frac{i}{z+i}\right)^n$$

which gives the inequality we are looking for :

$$\frac{|i|}{|z+i|}<1=>|z+i|>1$$

Finally, this leads to the Laurent Series:

$$f(z)=\sum_{n=0}^\infty\left(\frac{z+i}{1+i}\right)^n-\sum_{n=0}^\infty\left(\frac{i}{z+i}\right)^n$$

The left part is the analytic part , and the right one is the principal part except for the n=0 term. Is everything alright with my solution? Should I provide more details if I'm asked in a test?

• Check your final answer again. Some factors are missing. Otherwise looks good – Lozenges Jun 28 '18 at 13:57

In fact, since $f(-i)\neq 0$, $f$ is analytic in a neighborhood of $z=-i$. A way to calculate the expansion is \begin{align} \frac{1}{z(z-1)} = \frac{1}{z-1}-\frac{1}{z} &= -\frac{1}{1+i}\frac{1}{1-(z+i)/(1+i)}+\frac{1}{i}\frac{1}{1-(z+i)/i}\\ &=-\frac{1}{1+i}\sum_{n=0}^\infty \left(\frac{z+i}{1+i}\right)^n- i \sum_{m=0}^\infty\left(\frac{z+i}{i}\right)^m\\ &=\sum_{n=0}^\infty \left(\frac{1}{i^{n+1}}-\frac{1}{(1+i)^{n+1}}\right)(z+i)^n \end{align}

The expansion is valid in $B(-i,1)$.

Edit: I didn't read that we are in $1<|z+i|<\sqrt{2}$. Just consider that the coefficients of the expansion are given by de formula $$a_n = \frac{1}{2\pi i}\int_{|z+i|=1.2}\frac{dz}{z(z-1)(z+i)^{n+1}}$$ Let $g_n(z)=z^{-1}(z-1)^{-1}(z+i)^{-n-1}$. Then $g_n(z)$ is meropmorphic in $\mathbb{C}$ and in the annulus has one simple pole in $z=0$ and, if $n\geq 0$, a pole of order $n+1$ in $z=-i$. So $$\lim_{z\to 0} zg_n(z)=-(i)^{-n-1}=\begin{cases} i^{1-n} & n \neq -1\\ -1 & n = -1 \end{cases}$$ If $n\geq 0$ $$\lim_{z\to -i} \frac{\partial^n}{\partial z^n}(z+i)^{n+1}g_n(z) = \lim_{z\to -i} \frac{\partial^n}{\partial z^n}f(z) = \frac{1}{i^{n+1}}-\frac{1}{(1+i)^{n+1}}$$

So, using Residues theorem $$a_n = \begin{cases} -(i+1)^{-n-1} & n\geq 0\\ -1 & n = -1\\ i^{1-n} & n<-1 \end{cases}$$

• Doesn't a Laurent series have negative powers of $z-z_0$ where $z_0$ the point around which you expand ? Why aren't they here in this case? Is it because you are expanding for $|z+i|<1$? – John Katsantas Jun 28 '18 at 13:01
• @John: It will only have negative powers if the center is a pole of the function. Since $i$ is not a pole, you wonâ€™t have negative powers. – Clayton Jun 28 '18 at 13:12
• I thought that the region defines whether you will have negative powers or not. In this problem for $|z+i|<1$ the region contains no poles. But for $\sqrt2>|z+i|>1$ it does contain a pole(z=0) and it's not analytic. The same goes for $|z+i|>\sqrt2$. – John Katsantas Jun 28 '18 at 13:21
• In fact, if $a_1,\dotsc,a_k$ are the singularities of $f$, the expansion at $z=-i$ is valid only in $B(-i,r)$ where $r=\inf_k d(-i,a_k)$. In this case, the expansion is valid only in $B(-i,1)$. – Rafael Gonzalez Lopez Jun 28 '18 at 13:24
• @RafaelGonzalezLopez We are asked to find the Laurent series in an annulus $1<|z+i|<\sqrt{2}$ not in $B(-i,1)$ – Lozenges Jun 28 '18 at 13:45