z=0
A function f(z)f(z) has a removable singularity at z=z_0z=z0 if
- f(z)f(z) is not defined at z=z_0z=z0
- defining a value for f(z)f(z) at z=z_0z=z0 makes it analytic.
The function f(z) = sin(z)/(z cos z)f(z)=sin(z)zcosz is not defined at z=0z=0. However, its limit as z to 0z→0 exists and is 1.
It is easy to see that if we define
g(z) = f(z) " for " z ne 0 g(z)=f(z) for z≠0 and g(0) = 1g(0)=1
then the resulting function g(z)g(z) is analytic at z = 0z=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
