Laurent series: addition and multiplication of series

[libelec]

Find the Laurent Series around z=0 for f(z) = \frac{{2{z^2}}}{{2{z^2} - 5z + 2}} + \sin (\frac{3}{{{z^2}}}), so that the series absolutely converges in z = -i

The Attempt at a Solution

The singularities of f are z = 0, z = 1/2 and z=2. The Laurent series will converge absolutely in z = -i for the one whose ROC is 1/2 <|z|< 2.

First, I tried to separate the polynomial fraction in partial fractions:

\frac{{2{z^2}}}{{2{z^2} - 5z + 2}} = 2{z^2}\left( {\frac{1}{{(z - 2)(z - \frac{1}{2})}}} \right) = 2{z^2}\left( {\frac{2}{{3(z - 2)}} - \frac{2}{{3(z - \frac{1}{2})}}} \right).

I left the 2z² outside because it is its own Laurent series around z = 0 for all z, so I can work the partial fractions easier (I don't remember how to do it when the grades of both polynomials are the same).

Second, I find the series for each fraction and for the sine around z = 0 in the ring 1/2 <|z|< 2. For that, I will have to find the Laurant series of the fraction with (z - 1/2) around a vecinity of infinite. I will compare the fractions with the geometrical series and use a variable change for the sine:

a) \frac{2}{{3(z - 2)}} = \frac{{ - 2}}{{3(2 - z)}} = \frac{{ - 1}}{{3(1 - \frac{z}{2})}} = \frac{{ - 1}}{3}\left( {\frac{1}{{1 - \frac{z}{2}}}} \right) = \frac{{ - 1}}{3}\sum\limits_{n = 0}^\infty {\frac{{{z^n}}}{{{2^n}}}} ,\forall z/\left| z \right| &lt; 2

b)\frac{{ - 2}}{{3(z - \frac{1}{2})}} \to \frac{{ - 2}}{{3z(1 - \frac{1}{{2z}})}} = \frac{{ - 2}}{{3z}}\left( {\frac{1}{{1 - \frac{1}{{2z}}}}} \right) = \frac{{ - 2}}{{3z}}\sum\limits_{n = 0}^\infty {\frac{1}{{{2^n}{z^n}}}} ,\forall z/\left| z \right| &gt; \frac{1}{2}

c)Since \sin (u) = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{u^{2n + 1}}}}{{(2n + 1)!}}}, using the variable change u = 3/z², \sin (\frac{3}{{{z^2}}}) = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{3^{2n + 1}}}}{{(2n + 1)!{z^{4n + 2}}}}}, this for all z except z = 0.

Conclusion: Therefore I have that
f(z) = 2{z^2}\left( {\frac{{ - 1}}{3}\sum\limits_{n = 0}^\infty {\frac{{{z^n}}}{{{2^n}}} - \frac{2}{{3z}}\sum\limits_{n = 0}^\infty {\frac{1}{{{2^n}{z^n}}}} } } \right) + \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{3^{2n + 1}}}}{{(2n + 1)!{z^{4n + 2}}}}} ,\forall z/\frac{1}{2} &lt; \left| z \right| &lt; 2


Now, I can't find a way to get an unique expression for the series' coefficients a_n. This is what is troubling me.

I tried it this way: since \frac{2}{{3z}}\sum\limits_{n = 0}^\infty {\frac{1}{{{2^n}{z^n}}}} = \sum\limits_{n = 0}^\infty {\frac{1}{{{{3.2}^{n - 1}}{z^{n + 1}}}}} = \sum\limits_{n = 1}^\infty {\frac{1}{{{{3.2}^{n - 2}}{z^n}}}} = \sum\limits_{n = - \infty }^{ - 1} {\frac{{{2^{n - 2}}{z^n}}}{3}} (is this right?), then, I can add that series with the other one inside the parenthesis, because there's no single power of z in both series that repeats itself (one is a series of only positive powers and 0, the other a series of only negative powers). But then I would have an expression for a_n for n>0 and another one for n<0. That wouldn't change much after I multiply by 2z², since it multiplies both series and keeps the fact that they don't have powers of z in common.

As the matter of fact, the expression for a_n for n<0 will get more complicated after I add the series of the sine function.

What am I missing or doing wrong?

Thanks

 

[libelec]

Nobody?