Laurent expansion

Question

Question is to write Laurent expansion of $f(z)=\dfrac{1}{(z-2)(z-1)}$ in the annulus $1<|z|<2$ based at $z=0$

I am aware of the method of partial fractions and writing expansions for both $\dfrac{1}{z-1}$ and $\dfrac{1}{z-2}$..

I am rying to do this using Laurent expansion formula..

We have $$f(z)=\sum_{n=-\infty}^{\infty}a_n z^n$$ where $$a_n=\frac{1}{2\pi i}\int_{|z|=r}\frac{f(z)}{z^{n+1}}dz$$ for any $1<r<2$. Easiest case is when $n=-1$ then it is just $$a_{-1}=\frac{1}{2\pi i}\int_{|z|=r}\frac{1}{(z-1)(z-2)}dz=\frac{1}{2\pi i}\left(\int_{|z|=r}\frac{1}{z-2}-\int_{|z|=r}\frac{1}{z-1}\right)$$

Now, as $r<2$ we see that $\frac{1}{z-2}$ is analytic inside the circle of radius $r$.. so first integral is zero and we are left with second integral..

$$a_{-1}=\frac{1}{2\pi i}\int_{|z|=r}\frac{1}{(z-1)(z-2)}dz=-\frac{1}{2\pi i}\int_{|z|=r}\frac{1}{z-1}$$

I got stuck after this...

Answer 1

I think i got it...

For $1<r<2$, we have to compute $$a_n=\frac{1}{2\pi i}\int_{|z|=r}\frac{1}{(z-2)(z-1)z^{n+1}}dz$$

For $n>0$ we have $$a_{-n}=\frac{1}{2\pi i}\int_{|z|=r}\frac{z^{n-1}}{(z-2)(z-1)}dz =\frac{1}{2\pi i}\int_{|z|=r}\frac{z^{n-1}}{(z-2)}-\frac{z^{n-1}}{(z-1)}dz$$

As $r<2$, the function $\frac{z^{n-1}}{(z-2)}$ is analytic in the disk $\{|z|\leq r\}$ so first integral is zero. We then have

$$a_{-n}=\frac{1}{2\pi i}\int_{|z|=r}\frac{z^{n-1}}{(z-2)(z-1)}dz =-\frac{1}{2\pi i}\int_{|z|=r}\frac{z^{n-1}}{(z-1)}dz$$

The function $z^{n-1}$ is analytic so, by cauchy integral formula, we have $$a_{-n} =-\frac{1}{2\pi i}\int_{|z|=r}\frac{z^{n-1}}{(z-1)}dz=-z^{n-1}|_{z=1}=-1$$

So, for all $n>0$ we have $a_{-n}=-1$.

For $n>0$ we compute
$$a_n=\frac{1}{2\pi i}\int_{|z|=r}\frac{1}{(z-2)(z-1)z^{n+1}}dz$$

By induction, i could see that $$\frac{1}{(z-2)(z-1)z^{n+1}}=\frac{1}{2^{n+1}}\frac{1}{z-2}-\frac{1}{2^{n+1}}\frac{1}{z}-\cdots-\frac{1}{2}\frac{1}{z^{n+1}}-\frac{1}{z-1}+\frac{1}{z}+\cdots+\frac{1}{z^{n+1}}$$

As $\frac{1}{z-2}$ is analytic in the disk $\{|z|\leq r\}$ the integral is zero. Then, by all integrals are zero except $\frac{1}{z}$ and $\frac{1}{z-1}$.. By cauchy integral formula, $\frac{1}{z-1}$ integral is $1$.. We then have

$$a_n= \frac{1}{2\pi i}\int_{|z|=r}\left[\frac{1}{2^{n+1}}\frac{-1}{z}-\frac{1}{z-1}+\frac{1}{z}\right]=\frac{-1}{2^{n+1}}-1+1=\frac{-1}{2^{n+1}}$$

So, we have $a_n=\frac{-1}{2^{n+1}}$ for all $n\geq 0$ and $a_{-n}=-1$ for all $n>0$.

We then have $$f(z)=\cdots-\frac{1}{z^2}-\frac{1}{z}-\frac{1}{2}\left(1+\frac{z}{2}+\frac{z^2}{2^2}+\cdots\right)$$

Answer 2

For $\;1<r<2\;$ and using Cauchy's Integral Formula in the integral you gave:

$$a_{-1}=\frac1{2\pi i}\oint_{|z|=r}\frac{\frac1{z-2}}{z-1}dz=\left.\frac1{z-2}\right|_{z=1}=-1$$

Using partial fractions and around $\;z=1\;$ :

$$\frac1{z-2}-\frac1{z-1}=\frac1{1-(z-1)}-\frac1{z-1}=\sum_{k=0}^\infty(z-1)^k-\frac1{z-1}$$

as long as $\;|z-1|<1\;$ . From here, directly we get $\;a_{-1}=-1\;$, too