Calculate the principal part of $\pi\cot(\pi z)$

184 Views Asked by At

I calculated the principal part of $\pi\cot(\pi z)$ in $n\in\mathbb{Z}$ as $\dfrac1{z-n}$. Is this right?

1

There are 1 best solutions below

2
On BEST ANSWER

Yes your result is right. One has, for $n\in\mathbb{Z}$, as $z \to n$, $$ \begin{align} \lim_{z \to n}(z-n)\cdot \pi\cot(\pi z)&=\lim_{z \to n}\left((z-n)\cdot \pi\frac{\cos(\pi z)}{\sin(\pi z)}\right) \\\\&=\lim_{z \to n}\left((z-n)\cdot \pi\frac{\cos(\pi(z-n)+\pi n)}{\sin(\pi (z-n)+\pi n)}\right) \\\\&=\lim_{z \to n}\left((z-n)\cdot \pi\frac{(-1)^n\cos(\pi(z-n))}{(-1)^n\sin(\pi (z-n))}\right) \\\\&=\lim_{z \to n}\left(\frac{\pi (z-n)}{\sin(\pi (z-n))}\cdot\cos(\pi(z-n))\right) \\\\&=1\cdot1 \\\\&=1. \end{align} $$