1) Evaluate
the integral: òC (z + 1) dz / (2z +
i)
Solution:
A
simple (m=1) pole occurs at z = -i/2
Res
(f(z)) = (z – a) f(z) =
lim z ® -i/ 2 (2z + i) [(z + 1) / (2z + i)]
lim z ® -i/ 2 (2z + i) [(z + 1) / (2z + i)]
Res
(f(z)) = (z + 1)/ 2] z = -i/ 2 = (-i/2 + 1)/ 2 = ½
- i/4
(Remember if f(z) has the form p(z)/ q(z) and a first order pole exists at z = z o, then Res f(z) at z o =
lim z ® z o p(z) / q'(z) where q'(z) is the first derivative)
lim z ® z o p(z) / q'(z) where q'(z) is the first derivative)
Then:
òC (z + 1)
dz / (2z + i) = 2 pi (sum
of residues)
= 2 pi (½
- i/4) = pi - (pi)2/ 2 = pi + p/ 2
2)
Consider Example (2) and obtain the integral if we have
÷ z ÷ = 10 instead of ÷ z ÷ = 3/2
÷ z ÷ = 10 instead of ÷ z ÷ = 3/2
If
we now have ÷ z ÷ = 10
then both residues, i.e. for the simple and 2nd order poles can be
added since the singularities are now contained in the circle.
Then:
Sum
of residues = e/ 16 - 5 exp (-3) / 16
=
e/16 - 5e -3/ 16
So
that the re-evaluated integral is now:
2
pi (sum
of residues) = 2 pi (e/16
- 5e -3/ 16 )
= pi (e -
5e -3)/8
3)Evaluate
the integral: òC z dz
/ (z2 - 2z + 2)2
Solution, first rewrite:
f(z) =
z/ (z - 1 + i)2 (z - 1 -
i)2
In
the upper half plane we need to obtain Res(f(z) for z = 1+i
For
which we have a 2nd order pole.
Then:
lim z ® -1+i
d/ dz [(z - 1
- i)2 z/
(z - 1 + i)2 (z - 1 -
i)2]
= lim z ® -1+i
d/ dz [z/(z
- 1 + i)2 ]
And
from the product rule for differentiating, this yields:
2
+ a – 2z/ (z + a)3 for a = (-1
+ i)
lim z ® -1+i
d/ dz [z/(z
- 1 + i)2 ]
= (1 + i) + (-1 +i) – 2(1 + i)/ [(1 + i) + (-1
+i)] 3
= -2/ -8i
= 1/ 4i so that:
òC z dz / (z2 - 2z + 2)2 = 2 pi (1/ 4i) =
p/ 2
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