# How to know that the poles of sinz/{zcosz} at 0,+π/2 &-π/2 are simple poles?

May 19, 2018

The function has simple poles at $z = \pm \frac{\pi}{2}$

The singularity at $z = 0$ is not a pole - it is a removable singularity.

## z=0

A function $f \left(z\right)$ has a removable singularity at $z = {z}_{0}$ if

• $f \left(z\right)$ is not defined at $z = {z}_{0}$
• defining a value for $f \left(z\right)$ at $z = {z}_{0}$ makes it analytic.

The function $f \left(z\right) = \sin \frac{z}{z \cos z}$ is not defined at $z = 0$. However, its limit as $z \to 0$ exists and is 1.

It is easy to see that if we define

$g \left(z\right) = f \left(z\right) \text{ for } z \ne 0$ and $g \left(0\right) = 1$

then the resulting function $g \left(z\right)$ is analytic at $z = 0$. Indeed, it has the Taylor expansion

$g \left(z\right) = 1 + {z}^{2} / 3 + \frac{2 {z}^{4}}{15} + \setminus m a t h c a l \left\{O\right\} \left({z}^{5}\right)$

Thus $f \left(z\right) = \sin \frac{z}{z \cos z}$ has a removable singularity at $z = 0$

## z=pi/2

The function $f \left(z\right)$ is singular at $z = \frac{\pi}{2}$ (because $\cos z$ vanishes there). The limit

${\lim}_{z \to \frac{\pi}{2}} f \left(z\right)$

does not exist. However, the limit

${\lim}_{z \to \frac{\pi}{2}} \left(z - \frac{\pi}{2}\right) f \left(z\right) = {\lim}_{z \to \frac{\pi}{2}} \sin \frac{z}{z} \frac{z - \frac{\pi}{2}}{\cos} z$
$q \quad q \quad = {\lim}_{z \to \frac{\pi}{2}} \sin \frac{z}{z} {\lim}_{z \to \frac{\pi}{2}} \frac{z - \frac{\pi}{2}}{\cos} z$
$q \quad q \quad = \frac{2}{\pi} {\lim}_{z \to \frac{\pi}{2}} \frac{\frac{d}{\mathrm{dz}} \left(z - \frac{\pi}{2}\right)}{\frac{d}{\mathrm{dz}} \left(\cos z\right)}$
$q \quad q \quad = \frac{2}{\pi} {\lim}_{z \to \frac{\pi}{2}} \frac{1}{- \sin z} = - \frac{2}{\pi}$

exists, and thus $f \left(z\right)$ has a simple pole at $z = \frac{\pi}{2}$

A similar argument works for $z = - \frac{\pi}{2}$