f is holomorphic; if g(z) = $\pi$f(z)cot($\pi$z) show that $res_{z=z_0}$g(z) = f(n) for each n $\in$ $\Bbb Z$

109 Views Asked by At

Suppose f is holomorphic on an open set G $\subset$ $\Bbb C$ such that n $\in$ $\Bbb Z$ $\subset$ $\Bbb C$. If g(z) = $\pi$f(z)cot($\pi$z) show that $res_{z=z_0}$g(z) = f(n) for each n $\in$ $\Bbb Z$.

My initial thought is to use residue theorem and plug g(z) into the equation, but I am not sure how to handle the f(z)

1

There are 1 best solutions below

0
On

Hint:

This can be done by Laurent series expansion:

Laurent series about $n$: $$\pi \cot(\pi z)=\frac1{z-n}+O(1)$$ Taylor series about $n$: $$f(z)=f(n)+O(z)$$

When the two series are multiplied, the coefficient of the $(z-n)^{-1}$ term is $f(n)$.

As you know, the residue of a function at $k$ is the coefficient of the $(z-k)^{-1}$ term of its Laurent series about $k$.

You can make the argument more rigorous by manually doing a small loop contour integration about $n$, and then apply Cauchy’s integral formula.