- #1
polygamma
- 229
- 0
Show that the residue of $\cot^{n}(z)$ at $z=0$ is $\sin \left( \frac{n \pi}{2}\right)$, $n \in \mathbb{N}$.
The residue of $\cot^n(z)$ at $z=0$ is 1 if n is even, and 0 if n is odd. This can be proven using the Laurent series expansion of $\cot(z)$ around $z=0$.
To find the residue of $\cot^n(z)$ at $z=0$, you can use the formula $\text{Res}(f,c) = \frac{1}{2\pi i} \int_C f(z)dz$, where C is a small circle around $z=0$. You can then use the Laurent series expansion of $\cot(z)$ and the Cauchy Integral Formula to evaluate the integral.
The residue of $\cot^n(z)$ at $z=0$ is important in evaluating complex integrals involving $\cot^n(z)$. It can also be used in solving differential equations and in other areas of mathematics and physics.
No, the residue of $\cot^n(z)$ at $z=0$ can never be negative. This is because the residue is defined as the coefficient of the $\frac{1}{z}$ term in the Laurent series expansion, and this coefficient is always a non-negative integer.
The residue of $\cot^n(z)$ at $z=0$ is equal to the order of the pole at $z=0$ for $\cot(z)$. This means that if $z=0$ is a simple pole for $\cot(z)$, then the residue of $\cot^n(z)$ at $z=0$ is 1. If $z=0$ is a pole of higher order, the residue will be a higher power of $n$.