## z=0

A function #f(z)# has a removable singularity at #z=z_0# if

- #f(z)# is
**not** defined at #z=z_0#
- defining a value for #f(z)# at #z=z_0# makes it analytic.

The function #f(z) = sin(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(z) = f(z) " for " z ne 0 # **and** #g(0) = 1#

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

#g(z) = 1+z^2/3+{2z^4}/15+ \mathcal{O}(z^5)#

Thus #f(z) = sin(z)/(z cos z)# has a removable singularity at #z=0#

## z=pi/2

The function #f(z)# is singular at #z=pi/2# (because #cos z# vanishes there). The limit

#lim_{z to pi/2} f(z)#

does not exist. However, the limit

#lim_{z to pi/2} (z-pi/2)f(z) = lim_{z to pi/2} sin z/z {z-pi/2}/cos z#

#qquad qquad = lim_{z to pi/2}sin z/z lim_{z to pi/2}{z-pi/2}/cos z#

#qquad qquad = 2/pi lim_{z to pi/2}{d/dz(z-pi/2)}/{d/dz(cos z)}#

#qquadqquad = 2/pi lim_{z to pi/2}1/(-sin z) = -2/pi#

exists, and thus #f(z)# has a simple pole at #z=pi/2#

A similar argument works for #z=-pi/2#