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I need to classify all Singular points of the function into removable singularity, pole and essential singular point. Find the residue in them and order of every pole.

This is my function $f(z) = \dfrac 1{z(z-1)(z-2)}$. The singular points should be $0$, $1$ and $2$. Partial fraction gives me $f(z) = \frac 12 \left(\frac 1{z - 2} - \frac 2{z - 1} + \frac 1z\right)$ (sorry don't know to use math symbols here) I think the order of the poles is $1$ for all points but don't know how or if I can use residue theorem to find residue.

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up vote 3 down vote accepted

One could use the residue theorem to find the residues of those simple poles, but calculating the integral directly would probably be very difficult - the easiest way to calculate the relevant integrals would be to use the residue theorem.

It's easy to find the residues directly and the point of this exercise is to practice doing so. Either expand the terms you produced after partial fractions via geometric series to produce a Laurent series and read off the $a_{-1}$ coefficient, or use the short cut limit method for simple poles: $\displaystyle\text{Res}_{z=a} f(z) = \lim_{z\to a} (z-a) f(z) .$

I'll show you an example so you can get started. Using the limit method is certainly the quickest, but it is good practice to find Laurent series anyway. Say we want to find the residue at $z=0$. We want to write a Laurent series $\displaystyle f(z) = \sum_{n=-\infty}^{\infty} a_n z^n $ valid in some annulus around $z=0$. We know $$ f(z) = \frac{1}{2(z-2)} + \frac{1}{1-z} + \frac{1}{2z}.$$ The last term is already in the form we want. The middle term expands to $$ \frac{1}{1-z} = \sum_{n=0}^{\infty} z^n$$ for $|z|<1 $ and the first term can be written $$ \frac{-1/4}{1-z/2} = \frac{-1}{4} \sum_{n=0}^{\infty} \left( \frac{z}{2} \right)^n $$ for $|z/2|<1, \text{ or } |z|<2. $

Thus, in the annulus $0<|z|<1$ we have the Laurent series $$f(z) = \frac{1}{2z} + \sum_{n=0}^{\infty} \left( 1 - \frac{1}{2^{n+2}} \right) z^n.$$

We can read off that the residues at $z=0$ is $1/2$. Of course, we could have also noticed that the partial fraction decomposition left us with the form $$ f(z) = \frac{1}{2z} + \text{Function analytic at z=0} $$ and so the only negative powered term will be $\displaystyle \frac{1}{2z}$.

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