1. ## Laurent Series

hi im having a bit of trouble with laurnet series. im not really sure whats going on.

expand f(z) = 1/((z+1)(z+2)) in a laurnet series where the annulus contains the point z=7/2

i know that there are singularites at z=-1 and z=-2. and in the question i was given the hint to have the centre of the annulus at z=1. but im still not really sure what to do to expand the function.

any help would be great

thanks

2. Everything here is based on the binomial expansion $\tfrac1{1+x} = 1-x+x^2-x^3+\ldots$. This can be used in two ways to expand $\tfrac1{z+a}$, where a is a constant.

1. $\frac1{z+a} = \frac1{a(1+\frac za)} = \tfrac1a\bigl(1- \tfrac za + \bigl(\tfrac za\bigr)^2-\ldots\bigr)$ (expansion in positive powers of z);

2. $\frac1{z+a} = \frac1{z(1+\frac az)} = \tfrac1z\bigl(1- \tfrac az + \bigl(\tfrac az\bigr)^2-\ldots\bigr)$ (expansion in negative powers of z).

The expansion in positive powers of z converges when |z|<|a|. The expansion in negative powers of z converges when |z|>|a|.

To get the Laurent expansion for f(z) = 1/((z+1)(z+2)) that converges for z=7/2, use partial fractions and then, for each of the partial fraction terms, use whichever of the above series converges for z=7/2.

Edit. After reading the question more carefully (something it always pays to do ), I see that you want a series that converges in an annulus. So you want the singularities at z=-1 and z=-2 to be on the inner and outer circles of the annulus. You also want the point z=7/2 to be inside the annulus. So (as indicated in the hint) it would be a good idea to centre the annulus at the point z=1. Then the inner radius will be 2 (the distance from z=1 to z=-1), the outer radius will be 3 (the distance from z=1 to z=-2) and the distance from z=1 to z=7/2 is 5/2 which lies between 2 and 3.

The way to do the question is then to make a substitution w=z-1, so that the function becomes $\frac1{(z+1)(z+2)} = \frac1{(w+2)(w+3)}$. Then use the procedure that I described above to get the Laurent series for this function of w that converges when w=5/2. Finally, substitute back again to get a series in positive and negative powers of (z-1).