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Laurent series: $\sum_{n=0}^\infty a_nz^n +\sum_{n=0}^\infty b_nz^{-n} $

Questions: Find the coefficients $a_n,b_n$ of the laurent series for the function $f(z)=\frac{2+3z}{z^2+z^4}$ on the annulus $0< |z| < 1$.

Plan: We have a singularity of $z=0$ which is a pole of order $2$. Not really sure how to proceed after identifying the singularity.

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  • $\begingroup$ Use the geometric series expansion $1+q+q^2+\dots=1/(1-q)$ for $q=-z^2$, extract that $z^{-2}$ maybe first to have a "Taylor series". $\endgroup$
    – dan_fulea
    May 22 '19 at 17:08
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One way to go is to first note that $$z^2+z^4=z^2\left(z^2+1\right)=z^2(z-i)(z+i),$$ and that the numerator's degree is less than that of the denominator. Thus, there are constants $a,b,c,d$ such that $$\frac{2+3z}{z^2+z^4}=\frac{a}{z}+\frac{b}{z^2}+\frac{c}{z-i}+\frac{d}{z+i}.$$ I leave it to you to find them. (This is known as a partial fraction decomposition, if you want to look up more about it.)

Next, we could use the fact that for $z\ne 0,$ we have that $$\frac{1}{z+i}=\frac{1}{z}\cdot\frac{1}{1-\left(-\frac{i}{z}\right)}$$ and that $$\frac{1}{z+i}=\frac{1}{i}\cdot\frac{1}{1-\left(-\frac{z}{i}\right)},$$ then use the fact that for $|w|<1,$ we know that $$\frac{1}{1-w}=\sum_{n=0}^\infty w^n.\tag{$\heartsuit$}$$ In order to make $|z|<1,$ it is enough to make $|w|<1$ for one of $w=-\frac{i}{z}$ or $w=-\frac{z}{i}.$ (I'll leave it to you to determine which one.) That will let us rewrite $\frac{d}{z+i}$ as a Laurent series that is valid in the annulus $0<|z|<1,$ and we can similarly rewrite $\frac{c}{z-i}.$ Since $\frac{a}{z}+\frac{b}{z^2}$ is already a Laurent series that's valid in the given annulus, then we can simply add them together, and obtain the series $$\sum_{n=0}^\infty a_nz^n+\sum_{n=1}^\infty b_nz^{-n}$$ that we wanted. Note the slight difference between mine and yours. Can you see the reason for it?


The above approach is probably good for you to run through, for practice's sake. However, in this case, we can proceed in an even easier way! Note that $$f(z)=\frac{2+3z}{z^2}\cdot\frac{1}{1-\left(-z^2\right)},$$ and that $0<|z|<1$ if and only if $0<\left|z^2\right|<1,$ so we can use $(\heartsuit)$ to conclude that $$\frac{1}{1-\left(-z^2\right)}=\sum_{n=0}^\infty\left(-z^2\right)^n=\sum_{n=0}^\infty(-1)^nz^{2n}.$$ Hence, $$f(z)=\sum_{n=0}\frac{2+3z}{z^2}\cdot(-1)^nz^{2n}=\sum_{n=0}^\infty(-1)^n(2+3z)z^{2n-2}=\sum_{n=0}^\infty(-1)^n2z^{2n-2}+(-1)^n3z^{2n-1}.$$ Can you take it from there to figure out what the sequences $\left\{a_n\right\}_{n=0}^\infty$ and $\left\{b_n\right\}_{n=1}^\infty$ should be?

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Note that $$ \frac{2+3z}{z^2+z^4} = \frac{2}{z^2}\frac{1}{1+z^2}+ \frac{3}{z}\frac{1}{1+z^2}$$ and $$ \frac{1}{1+z^2}=\sum_{n=0}^\infty (-1)^n z^{2n} \qquad \text{for } 0<|z|<1$$

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