Question 7.34: Prove that cot z = 1/z + ∑n(1/z - nπ + 1/nπ) where the summa...
Prove that \cot z=\frac{1}{z}+\sum\limits_{n}\left(\frac{1}{z-n \pi}+\frac{1}{n \pi}\right) where the summation extends over n= \pm 1, \pm 2, \ldots
Consider the function
f(z)=\cot z-\frac{1}{z}=\frac{z \cos z-\sin z}{z \sin z}
Then f(z) has simple poles at z=n \pi, n= \pm 1, \pm 2, \pm 3, \ldots, and the residue at these poles is
\underset{z \rightarrow n \pi}{\lim}(z-n \pi)\left(\frac{z \cos z-\sin z}{z \sin z}\right)=\underset{z \rightarrow n \pi}{\lim}\left(\frac{z-n \pi}{\sin z}\right) \underset{z \rightarrow n \pi}{\lim}\left(\frac{z \cos z-\sin z}{z}\right)=1
At z=0, f(z) has a removable singularity since
\underset{z \rightarrow 0}{\lim}\left(\cot z-\frac{1}{z}\right)=\underset{z \rightarrow 0}{\lim}\left(\frac{z \cos z-\sin z}{z \sin z}\right)=0
by L’Hospital’s rule. Hence, we can define f(0)=0.
By Problem 7.110, it follows that f(z) is bounded on circles C_{N} having center at the origin and radius R_{N}=\left(N+\frac{1}{2}\right) \pi. Hence, by Problem 7.33,
\cot z-\frac{1}{z}=\sum\limits_{n}\left(\frac{1}{z-n \pi}+\frac{1}{n \pi}\right)
from which the required result follows.