Laurent series expansion of function $f(z)=\frac{1}{z^2-z}$ in powers of $z-i$ in region of convergence

Question

I need a little help with this one. This $z-i$ confuses me.

$$f(z)=\frac{1}{z^2-z}=\frac{1}{z(z-1)}=-\frac{1}{z}+\frac{1}{z-1}$$

So, singularities are $z=0$ and $z=1$. What regions do I have to consider when expanding in series?

And is this correct: if it was asked "in powers of $z$", in this particular case would we expand series in these areas $|z|<1$ and $|z|>1$?

There is nothing more in example text. Any help very appreciated.

Answer 1

Since you region of convergence is centered at $z=i$ and have singularities at $z=0,1$, therefore you need to consider the following regions:

1. $|z-i|<1$ (note:function is analytic in this region so you know the Laurent series will be simply ....)
2. $1<|z-i|<\sqrt{2}$
3. $|z-i|>\sqrt{2}$

This way you will be able to take care of both the singularities.

Now for the first region: Consider $$f(z)=-\frac{1}{z}+\frac{1}{z-1}=-\frac{1}{z-i+i}+\frac{1}{z-i+i-1}=-\frac{1}{i\left(1+\frac{z-i}{i}\right)}+\frac{1}{(i-1)\left(1+\frac{z-i}{i-1}\right)}.$$ Since we are in the first region, therefore $$\left|\frac{z-i}{i}\right| < 1 \text{ and } \left|\frac{z-i}{i-1}\right|< 1.$$ Thus we can use the following geometric series expansion to expand both terms. $$\frac{1}{1+w}=1-w+w^2-w^3+\dotsb \qquad (|w|<1),$$ and get $$f(z)=\frac{-1}{i}\sum_{k=0}^{\infty}(-1)^k\left(\frac{z-i}{i}\right)^k+\frac{1}{i-1}\sum_{k=0}^{\infty}(-1)^k\left(\frac{z-i}{i-1}\right)^k.$$ Similarly you can try the other two regions. For example for the second region you may want to start with $$f(z)=-\frac{1}{z-i+i}+\frac{1}{z-i+i-1}=-\frac{1}{(z-i)\left(1+\frac{i}{z-i}\right)}+\frac{1}{(i-1)\left(1+\frac{z-i}{i-1}\right)}.$$ Observe that since we are in the second region, therefore $$\left|\frac{i}{z-i}\right| < 1 \text{ and } \left|\frac{z-i}{i-1}\right|< 1.$$ Hopefully you will be able to take it from here.