# Laurent Series

• Jun 17th 2007, 05:10 AM
moolimanj
Laurent Series
Hi all

I'm trying to get my head around the Laurent series for the following function :

f(z) = zsinh(1/z+1)

about -1 and trying to give the general term of the series for odd and even powers of (z+1).

Also, how would I write down a punctured open disk D, containing the circle C={z:|z+1|=1} of which f is represented on this series, and what would be the nature of teh singularity of f at -1.

Any help would be greatly appreciated.

Many thanks
• Jun 17th 2007, 07:29 AM
ThePerfectHacker
Is that 1/(z+1) or 1/z + 1
• Jun 17th 2007, 07:55 AM
moolimanj
its 1/(z+1)
• Jun 17th 2007, 09:43 AM
ThePerfectHacker
Quote:

Originally Posted by moolimanj
f(z) = zsinh(1/z+1)

$\displaystyle f(z) = z \cdot \sinh \left( \frac{1}{z+1} \right)$

First we know that,
$\displaystyle \sinh w = w + \frac{w^3}{3!}+\frac{w^5}{5!}+... = \sum_{n=0}^{\infty} \frac{w^{2n+1}}{(2n+1)!}$

So if, $\displaystyle w=\frac{1}{z+1}$ we have,

$\displaystyle \sinh \left( \frac{1}{z+1} \right) = (z+1)^{-1} + \frac{(z+1)^{-3}}{3!} + \frac{(z+1)^{-5}}{5!}+...$

Now think of $\displaystyle z$ as $\displaystyle (z+1) - 1$

Thus,
$\displaystyle [ (z+1) - 1 ] \left( (z+1)^{-1} + \frac{(z+1)^{-3}}{3!} + \frac{(z+1)^{-5}}{5!}+... \right)$

Factor,
$\displaystyle \left(1 + \frac{(z+1)^{-2}}{3!} + \frac{(z+1)^{-4}}{5!} + ... \right) - \left( (z+1)^{-1} + \frac{(z+1)^{-3}}{3!} + \frac{(z-1)^{-5}}{5!}+ ... \right)$

Combine,
$\displaystyle -(z+1)^{-1} + \frac{(z+1)^{-2}}{3!} - \frac{(z+1)^{-3}}{3!} + \frac{(z+1)^{-4}}{5!} - \frac{(z+1)^{-5}}{5!} + ...$

Now you can state the formula for the general term.