Let f(z) be analytic on and inside a closed contour C as shown
below, except for a finite number of isolated singularities: z= a1, a2, etc.
which are enclosed by C.
Then: ò C 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:
ò C f(z)
dz = 2 pi [a - 1 +
b - 1
+ c - 1 + …………………….]
That is, 2 pi times the sum of the residues at all the
singularities enclosed by C. To ensure this, one would respectively construct
circles C1, C2, C3 etc. as I have done with respective centers at a, b, c etc.
If we take care to do this properly then we can write:
ò C f(z) dz = òC1 f(z) dz + òC2 f(z) dz + òC3 f(z) dz + ..........
Where:
ò C f(z) dz = òC1 f(z) dz + òC2 f(z) dz + òC3 f(z) dz + ..........
Where:
ò C1 f(z)
dz = 2 pi a - 1
ò C2 f(z)
dz = 2 pi b - 1
ò C3 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: òC cot
(z) dz
f(z) = cot (z)
For which: ò C f(z)
dz = 2 pi c - 1
Re-write: f(z) = cot (z) = 1/ tan z
For which singularities occur at tan
z = 0
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
ò C cot
(z) dz
= 2 pi (1) = 2 pi
Example 2:
Evaluate the
integral: 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
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 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 )
Practice Problems:
1) Evaluate
the integral: ò C (z + 1)
dz / (2z + i)
2) Evaluate the integral: ò C z dz
/ (z2 - 2z + 2)2
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