We introduce the Laurent Series as follows:
f(z)
= å¥ n = -¥ c n (z – a) n + å¥ n =-¥ c -n / (z – a) n
where:
c n = 1/2 pi òC1 f(w) dw /(w – a) n+1 n = 0,1, 2…..
Note
this series is unique for a given annulus, i.e. the shaded region shown in the image.
1 / (w – z) = 1/ {w – a) – (z – a) = 1/ (w – a)[ 1/ 1 –(z –a)/(w – a)]
Which will be: - å¥ m = 0 (z –
a) m / (z – a) m+1
Which
is convergent by the ratio test. Then:
1/2
pi òC1 f(w) dw /(w – z)
= - 1/2 pi òC1 f(w) dw /(w – z)
= - å¥ m = 0
1 / (z – a) m+1 1/2
pi òC1 f(w) (w – a) m
dw
Now,
replace the positive index m by –(n+1) and rewrite the preceding as:
1/2
pi òC1 f(w) dw /(w – z) =
å -¥ n = -1 (z – a) n 1/2 pi òC1 f(w) (w – a) n +1 dw
å -¥ n = -1 (z – a) n 1/2 pi òC1 f(w) (w – a) n +1 dw
Where
the integrals:
ò C2 f(w) dw /(w – a) n +1 n = -1,-2, -3
òC1 f(w) dw /(w – a) n+1 n = 0,1, 2….
Can
also be evaluated over a common circle C, concentric with C1 and C2 and lying
just within the annulus: R1 < R <
R2. To
prove uniqueness, assume an expansion:
å¥ n = -¥ c n (z – a) n
exists
and is valid in the annulus R1 < ÷ z – a ÷ < R2
Now
choose some arbitrary integer k and
multiply both sides of the expression by
:
(z
– a) –k +1 and integrate
around a circle C about z = a lying inside the annulus.. Then:
òC f(z) dz /(z – a) k+1 å¥ n = -¥ c n òC f(z) dz /(z – a) k+1 - n
Now,
all integrals on the right side will vanish except for one, for which n = k and
whose value is 2 pi
. Therefore:
òC f(z) dz /(z – a) k+1 =
c k 2 pi
Note
that the part of the Laurent series consisting of positive powers of (z – a) is
called the regular part, This resembles
the Taylor series that we already saw – but it needs to be clarified that the
nth coefficient can’t be disassociated in general with any nth derivative f n (a) since the latter may not
exist. (In most applications f(z) is not analytic at z = a).
The
other part of the series, consisting of negative powers, is called the
principal part. Either part or both
may terminate or be identically zero. If the principal part is identically zero
then f(z) is analytic at z = a since the derivative exists and the Laurent
series is identical to the Taylor
series.
More intricacies – and
singularities:
Point
z = a is called a zero or root of the function f(z) if f(a) = 0. If then f(z)
is analytic at at z = a then the Taylor
series:
f(z)
= å¥ n = 0
c n (z – a) n
must
have c 0 = 0.
If c 1 ¹ 0, the point a is called a simple zero
(or a zero of order one). It could happen that c 1
and perhaps several other next coefficients vanish. Then let c m be the
next vanishing coefficient (unless f(z) = 0) then the zero is said to be of
order m. The order of a zero may be evaluated – without any knowledge of the Taylor series – by
calculating:
lim z® a
f(z) / (z – a) n
for
n = 1, 2, 3. The lowest value of n for which this limit doesn’t vanish is equal
to the order of the zero.
1)
f(z) remains bounded, i.e. ÷ f(z)÷ <B for a fixed B
2) f(z) is not bounded and ÷ f(z)÷ approaches infinity. Namely,
÷ f(z)÷ > M
for ÷ z – a ÷ < e
3) Neither
of the two cases above, in other words f(z) oscillates.
Examples:
1) f(z)
= 1/ z – 1 (isolated singularity at z = 1)
a) If we demand ÷ z ÷ < 1 we can obtain a Taylor expansion.
b) )
If we seek: 1 < ÷ z ÷ < 0
we obtain a Laurent expansion
2)
f(z)
= 1/ z (1 – z)
a)
If 0 < ÷ z ÷ < 1 then
we obtain a Laurent series
b) If 1 < ÷ z ÷ < 0 we also obtain a Laurent expansion
Problem
for the Math Maven
Consider
the function: f(z) = 1/ (z+ 1) ((z + 3)
a)
Find a Laurent series for: 1
< ÷ z ÷ < 3
b)
Find a Laurent series for 0 < ÷ z + 1 ÷ < 2
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