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#
