Page 751 - Advanced_Engineering_Mathematics o'neil

22.1 Singularities 731

EXAMPLE 22.4
The Laurent expansion of e 1/z about 0 is
∞
1 1
1/z
e =
n! z n
n=0
for z = 0. f has an essential singularity at 0, because infinitely many powers of 1/z appear in
this expansion.

A pole of order 1 is called a simple pole, and a pole of order 2 is a double pole.

We do not want to have to resort to a Laurent expansion to classify a singularity. For the rest
of this section, we will explore other ways to do this.

THEOREM 22.1 Condition for a Pole of Order m

Let f be differentiable in 0 < |z − z 0 | < R. Then f has a pole of order m at z 0 if and only if
m
lim(z − z 0 ) f (z)
z→z 0
exists and is finite and nonzero.
Proof We can understand how this condition arises by manipulating the Laurent expansion of
f (z) about z 0 :
∞

f (z) = c n (z − z 0 ) n
n=−∞
for 0 < |z − z 0 | < R.If f has a pole of order m at z 0 , then c −m = 0 and c −m−1 = c m−2 = ··· = 0, so
the Laurent about the z 0 expansion is
c −m c −m+1
f (z) = + + ··· .
(z − z 0 ) m (z − z 0 ) m−1
Then
2
m
(z − z 0 ) f (z) = c −m + c −m+1 (z − z 0 ) + c −m+2 (z − z 0 ) + ··· ,
so
lim(z − z 0 ) f (z) = c −m = 0.
m
z→z 0
We will omit the details of the proof of the converse.
5
To illustrate the idea, look again at Example 22.3 with g(z)=sin(z)/z and z 0 =0. Compute
sin(z) sin(z)
4
4
lim z g(z) = lim z = lim = 1,
z→0 z→0 z 5 z→0 z
which is a limit that can be seen by using the Maclaurin expansion of sin(z) about 0 to write
sin(z) z 2 z 4
= 1 − + − ··· .
z 3! 5!
5
Theorem 22.1 tells us that sin(z)/z has a pole of order 4 at 0, as we found in Example 22.3 by
examining the Laurent expansion.

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October 14, 2010 15:37 THM/NEIL Page-731 27410_22_ch22_p729-750