Question : Prove that each singular point of tan z is a simple pole.
Solution :
Singularities of tan z Since we can write \[tan\ z\ =\frac{\sin{z}}{\cos{z}}=\frac{p\left(z\right)}{q\left(z\right)}\] Singularities of tan z are the zeros of cos z \[\cos{z}=0\] \[\cos{z}=\cos{\frac{n\pi}{2}},\ n=\pm1,\pm3,\pm5\ldots.. \] Thus singularities of tan z are npi/2 for every non zero integer n. A singular point z0 is a simple pole of a function f(z) if the following limit exists and finite \[p\left(\frac{n\pi}{2}\right)=\sin{\frac{n\pi}{2}}=\pm1\neq0 \] And \[q^\prime\left(z\right)=-\sin{z}\] \[ q^{\prime\left(\frac{n\pi}{2}\right)}=-\sin{\frac{n\pi}{2}}=\pm1\neq0\] Therefore each singular point of tan z is a simple pole.