# Laurent Series - Complex Numbers

• Jun 13th 2007, 05:07 AM
moolimanj
Laurent Series - Complex Numbers
(a) Locate and classify the singularities (giving the order of any poles) of the function
f(z) = z/(1-e^z)

(b) Let f(z) = zsinh (1/(z+1))
(i) Find the laurent series of f about -1, giving teh general term of the series for odd and even powers of (z+1)
(ii) Write down a punctured open disk D, containing the circle C={z:|z+1|=1} on which f is represented by this series
(iii) State the nature of the singularity of f at -1
(iv) Evaluate

integral zsinh(1/(z+1))dz where C={z:|z+1|=1}
(c) Find the laurent series about 0 for the function

f(z)=7z/((2z+1)(z-3)) on the set {z:|z|.3, giving teh general term of the series
• Jun 13th 2007, 06:15 AM
ThePerfectHacker
Quote:

Originally Posted by moolimanj
(a) Locate and classify the singularities (giving the order of any poles) of the function
f(z) = z/(1-e^z)

$f(z) = - \frac{z}{e^z-1}$
We see that at $z=0$ we have a possible singulariy.
Introduce some power series to this problem,
$- \frac{z}{1+z + \frac{z^2}{2!}+\frac{z^3}{3!}+.... - 1} = \frac{z}{z\left( 1 +\frac{z}{2!}+\frac{z^2}{3!}+... \right)} = \frac{1}{1+\frac{z}{2!}+\frac{z^2}{3!}+... }$
Now as $z\to 0$ we see that $f(z) \to \frac{1}{1+0+0+...} = 1$.
Thus, $z=0$ is a "removable singularity".
• Jun 17th 2007, 05:15 AM
moolimanj
Thanks PerfectHacker
Thanks for all your help with my questions to date PerfectHacker.

You are the one!:D