In a series of 2013 posts I examined Laurent series and it is useful to revisit this part of complex analysis again.

Let the function f(z) ibe analytic inside and on the boundary of the ring-shaped region **R** (see diagram) bounded by two concentric circles C1 and C2 with center at a and respective radii r1 and r2 (r1 > r2) then for all z in R. Then:

f(z) = å

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

c

**= 1/2 pi**_{n}**ò**_{C1}^{ }f(w) dw /(w – a)^{n+1}n = 0,1, 2…..**= 1/2 pi**

_{- n }**ò**

_{C2}^{ }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. The integral can be treated by expanding 1 / (w – z) i.e.

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+}^{}

Which series is convergent by the ratio test. Then:

1/2 pi

**ò**_{C1}^{ }f(w) dw /(w – z) = - 1/2 pi**ò**_{C1}^{ }f(w) dw /(w – z)= - å

^{¥}_{m = 0}1 / (z – a)^{m+1 }^{ }1/2 pi**ò**_{C1}^{ }f(w) (w – a)^{m}dwNow, replace the positive index m by –(n+1) and rewrite the preceding as:

1/2 pi

**ò**_{C1}^{ }f(w) dw /(w – z) =å

^{-}

^{¥}

_{n}

_{ = -1}(z – a)

^{n}1/2 p

**ò**

_{C1}^{ }f(w) (w – a)

^{ }

^{n +1}dw

Where the integrals:

**ò**

_{C2}^{ }f(w) dw /(w – a)

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

**ò**

_{C1}^{ }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 a 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 thePoint 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

f(z) = å

^{¥}_{n = 0}c**(z – a)**_{n}^{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.

**(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:**

*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*f(z) = 1/ z (1 – z)*

a) If 0 <

**÷**z**÷**< 1 then we obtain a Laurent seriesb) If 1 <

**÷**z**÷**< 0 we also obtain a Laurent expansion__Problems for the Math Maven__

1) 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**2) The function: f(z) = -1/ (z – 1) (z – 2)

a) Rewrite it using the partial fraction form

__the two singular points__.

^{2}+ 9 z - 18 / z

^{3}- 9z

Find Laurent series for the convergence regions:

a) 0 <

**÷**z

**÷**

**< 3 and**

b)

**÷**z**÷****> 3**
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