1) Find
the residue for f(z) = 3 exp (z)/ z 4
We
know from the residue formula:
Res f(z) = c - 1
The
pole is of order m = 4 at z = 0, then:
Res f(z) = 1/ 3! lim z ® 0
d 3
/ dz3 {(z - 0) 4 × 3 exp (z) / z 4 }
Res
f(z) = 1/ 3! lim z ® 0
d 3
/ dz 3 { 3exp (z)}
Res
f(z) = 1/6 lim z ® 0
{ 3exp (z)} = 3/6 = ½
2)
Find Res f(z) for cos z/ z 5
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) }
The
pole is of order m = 5 at z = 0 so that:
Res f(z) = 1/ 4! lim z ® 0
d 4
/ dz4 {(z - 0) 5 × cos (z) / z 5 }
Res
f(z) = 1/ 4! lim z ® 0
d 4
/ dz 4 { cos (z)}
Taking
the 4th derivative of cos z:
1)
d/ dz (cos z) = - sin z and 2) d/dz
(-sin z) = - cos z
3)
d/dz (- cos z) = - (-sin z) = sin z and
(4) d/dz (sin z) = cos z
3) Find
all the residues at those singular
points inside the circle ÷ z ÷ =
2
For: f(z) =
z 2 / ( z 4 - 1)
Rewrite
f(z):
z
2 / ( z 4 - 1)
= z 2 / ( z 2 - 1) ( z 2 + 1)
Look
at factors of the denominator:
(
z 2 + 1)
= (z + i) (z – i)
(
z 2 - 1) =
(z + 1) (z – 1)
Then
rewrite f(z) again:
z
2 / ( z 4 -
1) =
= z 2 / ( z - 1) ( z + 1) (z
+ i) (z – i)
The
order of the poles is simple, i.e. m = 1
and 4 singular points occur inside the
circle ÷ z ÷ = 2
z=
1, z = -1, z = i, z = -i
Look
at z = 1:
Res
(f(z)) = (z – a) f(z)
=
lim z ® 1 (z – 1) [z 2 / ( z - 1) ( z + 1) (z + i) (z – i)]
lim z ® 1 (z – 1) [z 2 / ( z - 1) ( z + 1) (z + i) (z – i)]
Res
(f(z)) = lim z ® 1
[z 2 / ( z + 1) (z
+ i) (z – i) ] =
z 2 / 4 z 3 ÷ z = 1 = 1 / 4(1) 3 = ¼
z 2 / 4 z 3 ÷ z = 1 = 1 / 4(1) 3 = ¼
Note:
by L’Hospital’s rule:
Res
[f, zo] = lim z ® z o 1 / ( z 4 - 1) =
lim z ® z o 1 / 4 z 3 = ( 1/ 4 zo 3 )
lim z ® z o 1 / 4 z 3 = ( 1/ 4 zo 3 )
Proceeding, for z = -1:
Res
(f(z)) = lim z ® -1 [z 2 / ( z - 1) (z
+ i) (z – i)] =
z 2 / 4 z 3 ÷ z =- 1 = (-1)2 / 4(-1) 3 = - ¼
z 2 / 4 z 3 ÷ z =- 1 = (-1)2 / 4(-1) 3 = - ¼
Similarly,
for z = -i:
Res
(f(z)) = lim z ® -i [z 2 / ( z - 1) (z
+ 1) (z – i) ]
= z 2 / 4 z 3 ÷ z =- -i = (-i)
2 / 4(-i) 3 = -1/ 4i
= 1/ 4i = -i/ -4 = i/4
Similarly, for z = i:
Res
(f(z)) = lim z ® i [z 2 / ( z - 1) (z
+ 1) (z + i)
= z 2 / 4 z 3 ÷ z = i = (i)
2 / 4(i) 3 = -1/ -4i
= 1/ 4i = -i/ 4
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