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I want to compute the laurent series of

i) $(z-3)\sin(\frac{1}{z+1}); (z\ne -1)$ at $z_{0}=-1 $.

ii) $\displaystyle{\frac{z}{(z-1)(z-2)}}; ( z\ne 1,2) $ at $z_{0}=1$

and classify their singularities.

A Laurent series has the form: $\displaystyle{f(z)= \sum_{n=1}^{\infty}c_{n}\frac{1}{(z-z_{0})^{n}}}+\sum_{n=0}^{\infty} b_{n}(z-z_{0})^{n}$

my attempts :

i) $\displaystyle{\sin(z):= \sum_{n=0}^{\infty} \frac{(-1)^{n}(x+1)^{2n+1}}{(2n+1)!}}$ at $z_{0}=-1 $ so for $x=\frac{1}{z+1}$ it follows :

$\displaystyle{(z-3)(\sin(x)) = (z-3) \sum_{n=0}^{\infty} \frac{(-1)^{n}(\frac{1}{1+z}+1)^{2n+1}}{(2n+1)!}}$ then I get stuck .

ii) partial fraction decomposition leads to: $\displaystyle{\frac{2}{z-2}-\frac{1}{z-1}}$ I can write these as taylor series . It is : $\displaystyle{\frac{1}{1-x} = \sum_{n=0}^{\infty}(x-1)^{n}}$

So it follows: $\frac{2}{z-2}-\frac{1}{z-1} = 2\sum_{n=0}^{\infty}(z-4)^{n} + \sum_{n=0}^{\infty}(z-2)^{n}$

then I get stuck

What is the right path?

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You're series expansion of $\sin(z)$ isn't quite right. It should be $\sum_{n=0}^\infty \frac{(-1)^n z^{2n+1}}{(2n+1)!}$. Keep in mind that the expansion of $z-3$ about $z=-1$ is $z-3=-4+(z+1)$ Therefore,

$$ (z-3)\sin\left(\frac{1}{z+1}\right) = \left(-4+(z+1)\right)\sum_{n=0}^\infty \frac{(-1)^n }{(2n+1)! (z+1)^{2n+1}}$$

This is

$$ (z-3)\sin\left(\frac{1}{z+1}\right) = \sum_{n=0}^\infty \frac{4(-1)^{n+1} }{(2n+1)! (z+1)^{2n+1}} + \sum_{n=0}^\infty \frac{(-1)^n }{(2n+1)! (z+1)^{2n}}$$

For the second problem, you only want to expand $2/(z-2)$ not both.

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

$$\frac{z}{(z-1)(z-2)} = -2\sum_{n=0}^\infty (z-1)^n - \frac{1}{z-1}$$

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Expansion of $sin(z)$ at $z_{0}=-1$ is $\displaystyle{ \sum _{n=0}^{\infty}\frac{(-1)^{n}(z+1)^{n+1}}{(2n+1)!} }$. So with $\displaystyle{(z-3)sin(\frac{1}{z+1})=\sum_{n=0}^{\infty} \frac{4(-1)^{n+1}}{(2n+1)!(z+1)^{2n+1}}} + \sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)!(z+1)^{2n}}$ how do I continue from here to find the laurent series? How can I find out about the singularities? – VVV Nov 3 '11 at 20:58
That is not the expansion of $\sin(z)$ at $z_0=-1$. The expansion of $\sin(z)$ at $z_0=-1$ involves evaluating $\sin$ and $\cos$ at $-1$ and looks quite ugly. Your expansion is that of $\sin(z+1)$ about $z_0=-1$. – Bill Cook Nov 3 '11 at 21:02
The second problem: You don't need to expand $1/(z-1)$, it's already expanded! It's Laurent series (centered at $z_0=-1$) is $\cdots+0(z-1)^{-2}+1(z-1)^{-1}+0(z-1)^0+0(z-1)+\cdots$ – Bill Cook Nov 3 '11 at 21:04
We use the series expansion about $z_0=0$ for $\sin(z)$ to find the series for $\sin(1/(z+1))$ about $z_0=-1$. The substitution changing $z$ to $1/(z+1)$ shifts the center of the expansion from $0$ to $-1$. – Bill Cook Nov 3 '11 at 21:25
You know you're done expanding into a series because you have written the function as a (infinite) sum of constants times powers of $z+1$. – Bill Cook Nov 3 '11 at 21:26

Is it correct that:

i) has a singularity that can be removed because it doesnt have any part of the form $\sum_{n=0}^{\infty}c_{n}(z-z_{0})^{-n}$ ?

ii) has a pole of order 1 as a singularity.

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