COMPLEX VARIABLES
whereais a finite, non-zero complex number. We note that if the above limit is
equal to zero, thenz=z 0 is a pole of order less thann,orf(z) is analytic there;
if the limit is infinite then the pole is of an order greater thann. It may also be
shown that iff(z) has a pole atz=z 0 ,then|f(z)|→∞asz→z 0 from any
direction in the Argand diagram.§If no finite value ofncan be found such that
(24.24) is satisfied, thenz=z 0 is called anessential singularity.
Find the singularities of the functions(i)f(z)=1
1 −z−
1
1+z, (ii)f(z)=tanhz.(i) If we writef(z)as
f(z)=1
1 −z−
1
1+z=
2 z
(1−z)(1 +z),
we see immediately from either (24.23) or (24.24) thatf(z) has poles of order 1 (orsimple
poles)atz=1andz=−1.
(ii) In this case we write
f(z)=tanhz=sinhz
coshz=
expz−exp(−z)
expz+exp(−z).
Thusf(z) has a singularity when expz=−exp(−z) or, equivalently, when
expz=exp[i(2n+1)π]exp(−z),wherenis any integer. Equating the arguments of the exponentials we findz=(n+^12 )πi,
for integern.
Furthermore, using l’Hopital’s rule (see chapter 4) we haveˆ
lim
z→(n+^12 )πi{
[z−(n+^12 )πi]sinhz
coshz}
= lim
z→(n+^12 )πi{
[z−(n+^12 )πi]coshz+sinhz
sinhz}
=1.
Therefore, from (24.24), each singularity is a simple pole.
Another type of singularity exists at points for which the value off(z)takesan indeterminate form such as 0/0 but limz→z 0 f(z) exists and is independent
of the direction from whichz 0 is approached. Such points are calledremovable
singularities.
Show thatf(z)=(sinz)/zhas a removable singularity atz=0.It is clear thatf(z) takes the indeterminate form 0/0atz= 0. However, by expanding
sinzas a power series inz, we find
f(z)=1
z(
z−z^3
3!+
z^5
5!−···
)
=1−
z^2
3!+
z^4
5!−···.
§Although perhaps intuitively obvious, this result really requires formal demonstration by analysis.