The Laurent Series of $\dfrac{e^z}{z^2-1}$

Question

The Laurent Series of $\dfrac{e^z}{z^2-1}$

At $z=1$

As we seek for powers of $z-1$, note that:

$$e^z=e\cdot e^{z-1}=e(1+(z-1)+\dfrac{(z-1)^2}{2!}+\dfrac{(z-1)^3}{3!}+...)$$

So: $f(z)=\dfrac{ee^{z-1}}{(1+(z-1))(z-1)}=e(\sum_{n=0}^{\infty}{(-1)^n(z-1)^n})(\dfrac{1}{z-1}+1+\dfrac{(z-1)}{2!}+\dfrac{(z-1)^2}{3!}+...)$

But I don't know what to do, becasue of the series multiplication, or I should have done it other way?

Answer 1

We can use Cauchy product (Cauchy product) to calculate the product series but first we need to correct the Laurent development of $\frac{1}{z^2-1}$: \begin{eqnarray} \frac{1}{z^2-1}&=&\frac{1}{z-1}.\frac{1}{z+1}\\ &=&\frac{1}{z-1}.\frac{1}{2+(z-1)}\\ &=&\frac{1}{z-1}.\frac{1}{2}\frac{1}{1-\frac{1-z}{2}}\\ &=& \frac{1}{2(z-1)}.\sum_{n\geq 0} \frac{(-1)^n}{2^n}(z-1)^n\\ &=& \sum_{n\geq 0} \frac{(-1)^n}{2^{n+1}}(z-1)^{n-1}\\ &=& \frac{1}{2}\frac{1}{z-1}+ \sum_{n\geq 0} \frac{(-1)^{n+1}}{2^{n+2}}(z-1)^{n} \end{eqnarray} and as you say the development of $e^z$ in $z=1$ is : $$ e^z=e \sum_{n\geq 0} \frac{1}{n!}(z-1)^{n} $$ so the development of the product is given by : \begin{eqnarray} f(z)&=&e^z.\frac{1}{z^2-1}\\ &=&e \left(\sum_{n\geq 0} \frac{1}{n!}(z-1)^{n}\right)\left( \frac{1}{2}\frac{1}{z-1}+ \sum_{n\geq 0} \frac{(-1)^{n+1}}{2^{n+2}}(z-1)^{n}\right) \\ &=& \frac{e}{2} \left(\sum_{n\geq 0} \frac{1}{n!}(z-1)^{n-1}\right)+ \sum_{n\geq 0}\left( \sum_{k=0}^n \frac{e}{k!}\frac{(-1)^{n-k+1}}{2^{n-k+2}} \right)(z-1)^n \\ &=& \frac{e}{2} \left(\sum_{n\geq -1} \frac{1}{(n+1)!}(z-1)^{n}\right)+ \sum_{n\geq 0}\left( \sum_{k=0}^n \frac{e}{k!}\frac{(-1)^{n-k+1}}{2^{n-k+2}} \right)(z-1)^n \\ &=& \frac{e}{2} \frac{1}{z-1}+ \sum_{n\geq 0}\left(\frac{e}{2} \frac{1}{(n+1)!}+\sum_{k=0}^n \frac{e}{k!}\frac{(-1)^{n-k+1}}{2^{n-k+2}} \right)(z-1)^n \\ &=& \frac{e}{2} \frac{1}{z-1}+ \sum_{n\geq 0}\left(\sum_{k=0}^{n+1} \frac{e}{k!}\frac{(-1)^{n-k+1}}{2^{n-k+2}} \right)(z-1)^n \end{eqnarray}

Answer 2

Note that $z_0=1$ is a simple pole, this mean that the Laurent around $z_0$ will have the following

$$ f(z)=\frac{a_{-1}}{z-1}+\sum_{\mathbb{N}}a_n(z-1)^n.$$

Now $$a_{-1}=\lim_{z\rightarrow 1}f(z)(z-1)=\frac{e^z}{z+1}=\frac{e}{2},$$so $$f(z)=\frac{e}{2(z-1)}+\sum_{\mathbb{N}}a_n(z-1)^n.$$

Now set $g(z)=f(z)(z-1)$; since $z_0=1$ is a simple pole, $g(z)$ is analytic in a neighborhood of $z_0$; in particular you can write $g(z)$ using Taylor serie:

$$g(z)=\sum_{\mathbb{N}}a_{n-1}(z-1)^n $$

and calculate the coefficients using the formula $a_{n-1}=\frac{g^{(n)}(1)}{n!}$. Clearly when $n=0$ you have $a_{-1}$ which is $\frac{e}{2}$.