To find the Laurent series of $\frac{1}{e^z-1}$, we first need to determine the singularities of the function. In this case, the only singularity is at $z=0$. Then, we can use the formula for the Laurent series: $\frac{1}{e^z-1}=\sum_{n=-\infty}^\infty a_nz^n$, where $a_n=\frac{1}{2\pi i}\oint_C \frac{f(z)}{(z-z_0)^{n+1}}dz$, and $C$ is a positively oriented contour around $z_0=0$.
A Taylor series is used to approximate a function around a specific point, while a Laurent series is used to represent a function in a region where it is not analytic, such as at a singularity. A Taylor series only includes non-negative powers of $z$, while a Laurent series includes both positive and negative powers.
No, the Laurent series of $\frac{1}{e^z-1}$ is only valid in the region where the function is not analytic, which is at $z=0$. Outside of this region, the series will not converge and cannot be used to evaluate the function.
The number of terms needed in a Laurent series to get a good approximation depends on the desired level of accuracy and the distance from the singularity. Generally, the more terms included, the better the approximation will be.
Yes, the Laurent series of $\frac{1}{e^z-1}$ can be used to evaluate integrals involving the function. This is because the Laurent series is a representation of the function itself, and can be used to calculate its values at any point within its region of convergence, including within an integral.