Recall we’d previously seen the function f(z) expressed as the Laurent series

f(z)
= å

^{¥}_{n = -}_{¥}c**(z – a)**_{n}^{n}+ å^{¥}_{n =-}_{¥}_{ }c**z – a)**_{-n / (}^{n}
We
also noted that no positive powers of z appear, only negative, i.e. 1/z is z

^{- 1}. So, in effect we could say that*the coefficients of the positive powers are zero*. It was also important to note here that the coefficient of :
1/ 1! z = 1/ z

is unity, so according to Laurent’s theorem we designate
that coefficient:

c

**= 1/2 pi**_{n}**ò**_{C}^{ }exp( 1/z) dz
where
C is any positively oriented simple closed contour around the origin. Further,
since n=
- 1 then, say for a function f(z) = exp(1/z):

c

**= 1/2 pi**_{- 1}**ò**_{C}^{ }exp( 1/z) dz and:**ò**_{C}^{ }exp( 1/z) dz = 2 piAnd in this way, one can actually compute what is called “

*the residue*” from a term of the Laurent series. The “

*residue theorem*” (due to Cauchy) is as follows:

**Let f(z) be analytic on and inside a closed contour C (see diagram) except for a finite number of isolated singularities at z = a1, a2…..etc., which are enclosed by C. Then:**

**ò**

_{C}^{ }f(z) dz = 2 pi å^{n}_{k = 1}Res f (a_{k})Alternately, Res f (a

**) = 1/2 pi**

_{k}**ò**

_{C}^{ }f(z) dz

Thus,
if we have some function f(z) centered at a we know we have a Laurent
expansion:

f(z)
= = å

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

_{- 1}
(There
is no need to compute the actual integral)

If f(z) is analytic at z = a the point z = a is then called a regular point and Res f(a) = 0. If, however, z = a is an isolated singularity, then the residue may or may not = 0.

Things
can be made even more straightforward by deriving a basic formula for computing
the residue.

Assume f(z) has a pole of

**order m**then the Laurent series of

f(z) is:

f(z)
= a

**/ (z – a)**_{- m}^{m }+ a**/ (z – a)**_{– m+ 1}^{m - 1 }+
a

**/ (z – a) + a**_{- 1}**+ a**_{0}**(z – a) + + a**_{1}**(z – a)**_{2}^{2}+ ……
Now,
multiply through by (z – a)

^{m }:
(z – a)

……. a

^{m }f(z) = a**+ a**_{- m}**(z – a) +**_{– m+ 1}……. a

**/ (z – a)**_{- 1}^{m - 1 }+ ….Which represents a

__about z= a of the analytic function, i.e. on the left. Now,__Taylor
series

*differentiate both sides*(m – 1) times with respect to z and you get:

d

^{ m - 1 }/ dz^{m - 1 }{(z – a)^{m }f(z) } = (m – 1)! a**+ m (m – 1) ……2 a**_{- 1}**(z – a) + …**_{2}
Then, let z ® a:

lim

_{ z }**d**_{® a}^{ m - 1 }/ dz^{m - 1 }{(z – a)^{m }f(z) } = (m – 1)! a_{- 1}
Where
bear in mind that a

**= c**_{- 1}**seen earlier.**_{- 1}
More
generally, we may write:

c

**= 1/ (m – 1)! lim**_{- 1}_{ z }**d**_{® a}^{ m - 1 }/ dz^{m - 1 }{(z – a**)**_{k}^{m }f(z) }
Examples:

1)Find
the residue for f(z) = 4

**/****(**1 - z)We have a simple pole (m=1) at z = 1.

Then
write:

Res
(f(z)) = (z – a) f(z)
= lim

= 4/ -1 = -4

_{ z }**(z – 1) [4**_{® 1}**/****(**1 - z)]= 4/ -1 = -4

(2) Find the residue for f(z)
= 1/ 1 – z

Again,
we have a simple pole at z = 1, so:

Res
(f(z)) = (z – a)

3)

^{ }f(z) = lim_{ z }**(z – 1) [1**_{® 1}**/****(**1 - z)] = -1
We
have a fourth order pole (m = 4) at z = 0, and we know from the residue
formula:

Res f(z) = c

= 1/ (m – 1)! lim

_{- 1}= 1/ (m – 1)! lim

_{ z }**d**_{® a}^{ m - 1 }/ dz^{m - 1 }{(z – a**)**_{k}^{m }f(z) }
Then:

Res f(z) = 1/ 3! lim

_{ z }**d**_{® 0}^{ 4 - 1 }/ dz^{4 - 1 }{(z)^{4 }**×****exp (z) / z**^{4 }}
Res f(z) = 1/ 3! lim

_{ z }**d**_{® 0}^{ 3 }/ dz^{3 }{ exp (z)}
=
1/ 3! lim

_{ z }**8 exp(2z) = 8/ 3! exp (0) = 8 / 6 = 4/3**_{® 0}
(4)
Find the residue of f(z) = z

^{2}exp (z)/ (z – 3)^{2 }
At
z = 3

We
have a 2

^{nd}order pole, i.e. m = 2 so that;
Res
f(z) = lim

_{ z }**d**_{® 3}^{ }/ dz^{ }{(z – 3)^{2 }[z^{2}exp (z)/ (z – 3)^{2 }] }
Res
f(z) = lim

_{ z }**d**_{® 3}^{ }/ dz^{ }{z^{2}exp (z)}
=
lim

= 15 exp (3)

_{ z }**exp (z) [ z**_{® 3}^{2}+ 2z ] = 9 exp (3) + 6 exp (3)= 15 exp (3)

__Problems for Math Mavens__:

1)
Find the residue for f(z) = 3 exp
(z)/ z

^{4}^{}
Be
sure to specify the order of the poles before proceeding!

2)Find
Res f(z) for cos z/ z

^{5}
3)
Find

**all the residues**at those singular points inside the circle**÷**z**÷****= 2**
For: f(z) =
z

^{2}/ ( z^{4}- 1)
(Hint:
remember your complex numbers!)

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