We introduce the Laurent Series as follows:

f(z)
= å

^{¥}_{n = -}_{¥}c**(z – a)**_{n}^{n}+ å^{¥}_{n =-}_{¥}_{ }c**z – a)**_{-n / (}^{n}
where:

c

**= 1/2 pi**_{n}**ò**_{C}**1**^{ }f(w) dw /(w – a)^{n+1}n = 0,1, 2…..**= 1/2 pi**

_{- n }**ò**

_{C}**2**

^{ }f(w) dw /(w – a)

^{–n +1}n = -1,-2, -3…..

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

**ò**_{C}**1**^{ }f(w) dw /(w – z) = - 1/2 pi**ò**_{C}**1**^{ }f(w) dw /(w – z)
= - å

^{¥}_{m = 0}1 / (z – a)^{m+1 }1/2 pi**ò**_{C}**1**^{ }f(w) (w – a)^{m}dw
Now,
replace the positive index m by –(n+1) and rewrite the preceding as:

1/2
pi

å

**ò**_{C}1^{ }f(w) dw /(w – z) =å

^{-}^{¥}_{n = -1}(z – a)^{n}1/2 pi**ò**_{C}1^{ }f(w) (w – a)^{ n +1}dw
Where
the integrals:

**ò**

_{C}**2**

^{ }f(w) dw /(w – a)

^{n +1}n = -1,-2, -3

**ò**

_{C}1^{ }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**(z – a)**_{n}^{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:**ò**

_{C}^{ }f(z) dz /(z – a)

^{k+1 }å

^{¥}

_{n = -}

_{¥}c

_{n }**ò**

_{C}^{ }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:

**ò**

_{C}^{ }f(z) dz /(z – a)

^{k+1 }= c

**2 pi**

_{k}
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 Taylor
series.

__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**:**

__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**(z – a)**_{n}^{n}
must
have c Taylor series – by
calculating:

**= 0. If c**_{0}**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**_{1}**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**_{m}
lim

_{ z}**f(z)**_{® a}_{ }**/**(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.

**(or isolated singular point) at z = a. It’s customary to distinguish isolated singularities by the following types of behavior of f(z) as z ® a for an arbitrary function.**

*isolated singularity*
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 Taylor expansion.

**÷**z**÷****< 1 we can obtain a**
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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