Recall we’d previously seen the function f(z) expressed as the Laurent series
f(z)
= å¥ n = -¥ c n (z – a) n + å¥ n =-¥ c -n / (z – a) 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 n = 1/2 pi 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 = 1/2 pi òC exp( 1/z)
dz and: òC exp( 1/z)
dz = 2 pi
And 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:
òC f(z) dz
= 2 pi ån k = 1 Res f (a k)
Alternately, Res f (a k) = 1/2 pi òC 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 n (z – a) n + å¥ n =-¥ c - n / (z – a) n
(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 - m / (z – a) m +
a – m+ 1 / (z – a) m - 1 +
a - 1 / (z – a) + a 0 + a 1
(z – a) + + a 2 (z – a)2
+ ……
Now,
multiply through by (z – a) m :
(z – a) m f(z) =
a - m
+ a – m+ 1 (z – a)
+
……. a - 1 / (z – a) m - 1 + ….
……. a - 1 / (z – a) m - 1 + ….
Which represents a
d
m - 1 / dz m - 1 {(z – a) m f(z) } =
(m – 1)! a - 1 + m (m
– 1) ……2 a 2 (z – a) + …
Then, let z ® a:
lim z ® a
d m - 1
/ dz m - 1 {(z – a) m f(z) } = (m – 1)! a - 1
Where
bear in mind that a - 1
= c - 1 seen earlier.
More
generally, we may write:
c - 1 =
1/ (m – 1)! lim z ® a
d 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 z ® 1 (z – 1)
[4 / (1 - z)]
= 4/ -1 = -4
= 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) f(z)
= lim z ® 1 (z – 1)
[1 / (1 - z)] =
-1
3)
We
have a fourth order pole (m = 4) at z = 0, and we know from the residue
formula:
Res f(z) = c - 1
= 1/ (m – 1)! lim z ® a d m - 1 / dz m - 1 {(z – a k) m f(z) }
= 1/ (m – 1)! lim z ® a d m - 1 / dz m - 1 {(z – a k) m f(z) }
Then:
Res f(z) = 1/ 3! lim z ® 0
d 4 - 1
/ dz 4 - 1 {(z) 4 × exp (z) / z 4 }
Res f(z) = 1/ 3! lim z ® 0
d 3
/ dz 3 { exp (z)}
=
1/ 3! lim z ® 0
8 exp(2z) = 8/ 3! exp (0) = 8 / 6 =
4/3
(4)
Find the residue of f(z) = z 2
exp (z)/ (z – 3) 2
At
z = 3
We
have a 2nd order pole, i.e. m = 2 so that;
Res
f(z) = lim z ® 3
d / dz {(z – 3) 2
[z 2 exp (z)/ (z – 3) 2
] }
Res
f(z) = lim z ® 3 d / dz {z 2 exp (z)}
=
lim z ® 3 exp (z) [ z2 + 2z ] = 9 exp (3) +
6 exp (3)
= 15 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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