eLaurentExpansion

Example Find Laurent series about the indicated singularity for the following function.

$\displaystyle f(z) = \frac{e^z}{z(1-z)}, a = 1$

Answer We let $z - 1 = t$ . Then $z = 1+t$ and $f(z) = \frac{e^{1+t}}{(1+t)(-t)}$ .

Now

$\displaystyle e^{1+t} =　 1 + 1+t + \frac{(1+t)^2}{2} + \frac{(1+t)^3}{3!} + \cdots$

Thus,

$\displaystyle \frac{e^{1+t}}{(1+t)(-t)} = \frac{1 + 1+t + \frac{(1+t)^2}{2} + \frac{(1+t)^3}{3!} + \cdots}{(1+t)(-t)}$

We note that for $\vert z\vert = \vert 1+t\vert < 1$ , we can expand $\frac{1}{1-(1+t)}$ .

$\displaystyle \frac{1}{-t} = \frac{1}{1-(1+t)} = 1 + (1+t) + (1+t)^2 + (1+t)^3 + \cdots$

This way, we can find Laurent series of $f(z)$

$\displaystyle f(z)$	$\displaystyle =$	$\displaystyle \frac{e^{1+t}}{(1+t)(-t)} = (\frac{1}{1+t} + 1 + \frac{1+t}{2} + \frac{(1+t)^2}{3!} + \cdots)(1 + (1+t) + (1+t)^2 + (1+t)^3 + \cdots)$
	$\displaystyle =$	$\displaystyle \frac{1}{1+t} + 2 + \frac{5}{2}(1+t) + \frac{8}{3}(1+t)^2 + \cdots$
	$\displaystyle =$	$\displaystyle \frac{1}{z} + 2 + \frac{5}{2}z + \frac{8}{3}z^2 + \cdots$

Note that $z = 0$ is simple pole for all $z \neq 1$ .