Recall that we saw the “residue theorem” (due to Cauchy)i.e. 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.
òC f(z) dz
= 2 pi ån k = 1 Res f (a k)
We now want to elaborate this a bit more by reference to the diagram shown. In this case we consider the function f(z) is analytic inside and ON the simple closed curve C except at a finite number of specified points: a, b, c, etc. at which there exist residues: a - 1 , b - 1 , c - 1 , etc.
In which case we can write:
òC f(z) dz
= 2 pi [a - 1 +
b - 1
+ c - 1 + …………………….]
ò C f(z) dz = ò C1 f(z) dz + ò C2 f(z) dz + ò C3 f(z) dz + ..........
Where:
òC1 f(z)
dz = 2 pi a - 1
òC2 f(z)
dz = 2 pi b - 1
òC3 f(z)
dz = 2 pi c - 1
So
that:
òC f(z) dz
= 2 pi [a - 1 +
b - 1
+ c - 1 + ..] =
2 pi (sum of residues)
Example
1:
Evaluate the integral:
ò C cot (z)
dz
f(z)
= cot (z)
For
which: ò C f(z)
dz = 2 pi c - 1
Or:
o, + p, + 2p,+ 3p
etc.
Then
Res f(z) = 1/ sec2 z ÷ z = + n p
=
1/ (1/ cos2 z)
=
cos2 z÷ z = + n p
=
cos2 (np)
And
: cos2 (np) = 1
at z = (2n + 1) p)/ 2
Therefore:
c - 1 =
1, and
òC cot (z)
dz = 2 pi (1) = 2 pi
Example
2:
ò C exp (z)
dz / (z – 1) (z + 3)2
Where
C is given by ÷ z ÷ = 3/2
Solution:
Take the residue at the simple pole (z = 1) such that:
lim z ® 1
[ (z – 1)
exp (z) / ( z - 1) (z
+ 3)2 ] =
exp(1)/ 16 = e/ 16
The
residue at the 2nd order pole (z = -3) is:
lim z ® -3
d/ dz [(z + 3)2 exp (z)
/ ( z - 1) (z + 3)2 ] =
lim z ® -3
[ (z – 1)
exp (z) - exp(z) / (z – 1 )2
]
= - 5 exp (-3) / 16
The
integral is therefore:
ò C exp (z)
dz / (z – 1) (z + 3)2 = 2 pi a - 1
= 2 pi (e/
16)
(We
do not add the 2nd residue because it lies beyond the circle ÷ z ÷ = 3/2 )
Problems for Math
Mavens:
2)
Consider Example (2) and obtain the integral if we have ÷ z ÷ = 10
instead of
÷ z ÷ = 3/2
3) Evaluate the integral: ò C z dz / (z2 - 2z + 2)2
in the upper half z-plane
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