LEIF MEJLBRO
COMPLEX FUNCTIONS EXAMPLES
C‐6
CALCULUS OF RESIDUES
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Leif Mejlbro
Complex Functions Examples c-6
Calculus of Residues
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Complex Functions Examples c-6 – Calculus of Residues
© 2008 Leif Mejlbro & Ventus Publishing ApS
ISBN 978-87-7681-391-8
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Contents
Complex Funktions Examples c-6
Contents
Introduction
5
1
Rules of computation of residues
6
2
Residues in nite singularities
9
3
Line integrals computed by means of residues
33
4
The residuum at ∞
57
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4
Introduction
Complex Funktions Examples c-6
Introduction
This is the sixth book containing examples from the Theory of Complex Functions. In this volume we
shall consider the rules of calculations or residues, both in finite singularities and in ∞. The theory
heavily relies on the Laurent series from the fifth book in this series. The applications of the calculus
of residues are given in the seventh book.
Even if I have tried to be careful about this text, it is impossible to avoid errors, in particular in the
first edition. It is my hope that the reader will show some understanding of my situation.
Leif Mejlbro
15th June 2008
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5
Complex Funktions Examples c-6
1
Rules of computation of residues
Rules of computation of residues
We refer in general to the following rules of computation of residues:
Definition of a residuum. Assume that f (z) is an analytic function defined in a neighbourhood of
z0 ∈ C (not necessarily at z0 itself ) with the Laurent series expansion
+∞
an z n ,
f (z) =
n=−∞
0 < |z| < r.
We define the residuum, or residue, of f (z) (more correctly of the complex differential form f (z) dz)
as the coefficient of 1/z in the Laurent series, i.e.
res (f (z) dz; z0 ) = res (f (z); z0 ) :=
1
2πi
Γ
f (z) dz = a−1 ,
where Γ denotes any simple closed curve, which surrounds z0 in positive sense, and where there is no
other singularity of f (z) inside and on the curve Γ.
Rule I. If z0 ∈ C is a pole of order ≤ q, where q ∈ N, of the analytic function f (z), then
dq−1
1
lim
(q − 1)! z→z0 dz q−1
(z − z0 )
q−1
f (z) .
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6
Complex Funktions Examples c-6
Rules of computation of residues
An important special case of rule I is
Rule Ia. If z0 is a simple pole or a removable singularity of the analytic function f (z), then
res (f ; z0 ) = lim (z − z0 ) f (z).
z→z0
Rule II. If A(z) and B(z) are analytic in a neighbourhood of z0 , and B(z) has a zero of first order
at z0 , then the residuum of the quotient f (z) := A(z)/B(z) is given by
res (f (z); z0 ) = res
A(z)
; z0
B(z)
=
A (z0 )
.
B ′ (z0 )
We also have the following generalization of Rule II, which however is only rarely used, because it
usual implies some heavy calculations:
Rule III. Assume that A(z) and B(z) are both analytic in a neighbourhood of z 0 , and assume that
B(z) has a zero of second order. Then the residuum of the quotient f (z) = A(z)/B(z) at z 0 it given
by
res (f (z); z0 ) = res
A(z)
; z0
B(z)
=
6 A′ (z0 ) B ′′ (z0 ) − 2 A (z0 ) B (3) (z0 )
3 {B ′′ (z0 )}
2
.
The complicated structure of Rule III above indicates why it should only rarely be applied.
Definition of the residuum at ∞. Assume that f (z) is analytic in the set |z| > R, so f (z) has
a Laurent series expansion
+∞
an z n .
f (z) =
n=−∞
We define the residuum at ∞ as
res(f (z) dz; ∞) := −a−1 ,
where one should notice the change of sign.
Rule IV. Assume that f (z) has a zero at ∞. Then
res(f dz; ∞) = − lim z f (z).
x→∞
Rule V. Assume that f (z) is analytic for |z| > R.Then
res(f (z) dz; ∞) = −res
1
f
w2
1
w
dw; 0 .
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7
Complex Funktions Examples c-6
Rules of computation of residues
This may be expressed in the following way: If we change the variable in the Laurent series expansion
above by z = 1/w, then the singularity z0 = ∞ is mapped into w0 = 0. Since
−
1
dw = d
w2
1
w
(= dz),
it follows by this change of variable that we have as a differential form
res(f (z) dz; ∞) = res f
1
w
d
1
w
; w0 = 0 ,
which shows that it is the complex differential form, which is connected with the residues.
Cauchy’s residue theorem. Assume that f (z) is analytic in an open domain Ω ⊆ C, and let Γ
be a simple, closed curve in Ω, run through in its positive direction, such that there are only a finite
number of singularities {z1 , . . . , zk } of f (z) inside the curve, i.e. to the left of the curve seen in its
direction. Then
k
1
2πi
res (f (z); zn ) .
f (z) dz =
Γ
n=1
Special case of Cauchy’s residue theorem. Assume that f (z) is analytic in Ω = C\{z 1 , . . . , zk },
i.e. f (z) has only a finite number of singularities in C. then
k
n=1
res (f (z); zn ) + res(f (z); ∞) = 0,
i.e. the sum of the residues is 0.
Finally, it should be mentioned that since functions like
1
,
sin z
1
,
cos z
tan z,
cot z,
1
,
sinh z
1
,
cosh z
tanh z,
coth z,
etc., does not have ∞ as an isolated singularity, none of these functions has a residuum at ∞.
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8
Complex Funktions Examples c-6
2
Residues in finite singularities
Residues in finite singularities
Example 2.1 Find the residuum of the function f (z) =
Then compute
1
, z = 0, 1, at the point 0.
z 2 (z − 1)
dz
.
− 1)
z 2 (z
|z|= 12
We expand f (z) into a Laurent series in the annulus 0 < |z| < 1, i.e. in a neighbourhood of z 0 = 0.
Then
f (z) =
1
1
1
1
=− 2 ·
=− 2
z 1−z
z
z 2 (z − 1)
+∞
n=0
zn = −
1
1
− − 1 − · · · − zn − · · · .
z2
z
The residuum is a−1 of this expansion, so it follows immediately that
res
1
,0
z 2 (z − 1)
= a−1 = −1.
Then
|z|= 12
dz
= 2πi res
z 2 (z − 1)
1
,0
z 2 (z − 1)
= −2π i.
Example 2.2 Find the residuum of the function f (z) =
1
, z = 0, 1, in the point 0.
z 2 n (z 2 − 1)
The function can be considered as a function in w = z 2 , so the Laurent series expansion from z0 = 0
only contains even exponents. In particular, a−1 = 0, hence
res]
1
z 2 (z 2 − 1)
= a−1 = 0,
and we do not have to find the explicit Laurent series in this case.
Example 2.3 Find the residuum of the function f (z) =
sin2 z
, z = 0, at the point z0 = 0.
z5
The numerator sin2 z has a zero of order 2, and the denominator z 5 has a zero of order 5, hence
sin2 z
f (z) =
has a pole of order 3 at z0 = 0.
z5
If we choose q = 3 in Rule I, we get the following expression,
rex
sin2 z
;0
z5
=
d2
1
lim 2
2! z→0 dz
sin2 z
z2
,
which will give us some unpleasant computations.
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9
Complex Funktions Examples c-6
Residues in finite singularities
Then note that Rule I gives us the possibility to choose a larger q, which here is to our advantage. In
fact, if we choose q = 5 in Rule I, then
sin2 z
;0
z5
rex
=
d3
d4
1
1
1
1
lim 23 {− cos 2z} = − .
lim 3 {sin 2z} =
lim 4 sin2 z =
24 z→0
3
24 z→0 dz
4! z→0 dz
Example 2.4 Find the residues at z = 0 of the following functions:
z2 + 1
,
z
(a)
(b)
z 2 + 3z − 5
.
z3
(a) It follows from
z2 + 1
1
= + z,
z
z
that
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10
Complex Funktions Examples c-6
Residues in finite singularities
(b) It follows from
z 2 + 3z − 5
1
3
5
= + 2 − 3,
z3
z
z
z
that
res(f ; 0) = a−1 = 1.
Example 2.5 Find the residues at z = 0 of the following functions:
(a)
ez
,
z
(b)
ez
,
z2
(z)
sin z
.
z4
(a) Here, z = 0 is a simple pole, hence by Rule I,
res(f ; 0) = lim ez = 1.
z→0
(b) Here, z = 0 is a double pole, hence by Rule I,
res(f ; 0) =
d z
1
e = lim ez = 1.
lim
z→0
1! z→0 dz
(c) We get by a series expansion of the numerator sin z that
sin z
1
= 4
z4
z
z−
z5
z3
− ···
+
5!
3!
=
1
1 1
1
− · +
· z + ··· .
z3
6 z
120
Hence
res
sin z
;0
z4
1
= a−1 = − .
6
Alternatively we apply Rule I, considering 0 as a pole of at most order 4 (the order is in fact
3 < 4):
res
sin z
;0
z4
=
d3
1
1
1
lim 3 sin z = lim {− cos z} = − .
z→0
z→0
dz
6
3!
6
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11
Complex Funktions Examples c-6
Residues in finite singularities
Example 2.6 Find the residues at z = 0 of the following functions:
(a)
sin z
,
z5
(a) Here,
res
Log(1 + z)
.
z2
(b)
sin z
is an even function, so
z5
sin z
;0
z5
= a−1 = 0.
Alternatively we prove this by a series expansion,
sin z
1
= 5
z5
z
z−
z5
z3
− ···
+
5!
3!
=
1
1 1
1
−
+
− ··· ,
z4
6 z2
120
from which we derive that
res
sin z
;0
z5
= a−1 = 0.
Alternatively we apply Rule I, because 0 is a pole of at most order 5 (the order is in fact 4):
res
sin z
;0
z5
=
1
1
d4
lim sin z = 0.
lim 4 sin z =
4! z→0
4! z→0 dz
(b) We have in a neighbourhood of 0 (exclusive 0 itself),
Log(1 + z)
1
= 2
z2
z
z−
z3
z2
+
− ···
2
3
=
1 1 z
− + − ··· ,
z
2 3
so
res
Log(1 + z)
;0
z2
= a−1 = 1.
Alternatively, z = 0 is a pole of at most order 2 (its order is 1), so by Rule I,
res
Log(1 + z)
;0
z2
=
d
1
1
lim
Log(1 + z) = lim
= 1.
z→0 1 + z
1! z→0 dz
Example 2.7 Find the residues of all singularities in C of
(a)
1
,
z(z − 1)
(b)
z4
z
,
+1
(c)
sin z
.
− z)
z 2 (π
(a) The function
f (z) =
1
z(z − 1)
has the simple poles 0 and 1. Then by Rule I:
res(f ; 0)
=
res(f ; 1)
=
1
= −1,
z→0 z − 1
lim z · f (z) = lim
z→0
1
= 1.
z→1 z
lim (z − 1)f (z) = lim
z→1
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12
Complex Funktions Examples c-6
Residues in finite singularities
(b) Here we have the four simple poles
exp i
π
,
4
exp i
3π
4
,
exp i
5π
4
,
exp i
7π
4
.
If we put
A(z) = z
B(z) = z 4 + 1,
and
and let z0 denote any of these simple poles, then z04 = −1 for all four of them, and we conclude by
Rule II that
res (f ; z0 ) =
z0
A (z0 )
1 1
z4
=
= · 4 · z02 = − 0 ,
3
′
4 z0
B (z0 )
4
4 z0
hence
π
1
exp i
4
2
i
=− ;
4
res f ; exp i
π
4
res f ; exp i
3π
4
=−
1
3π
exp i
4
2
=
res f ; exp i
5π
4
=−
5π
1
exp i
4
2
i
=− ;
4
res f ; exp i
7π
4
=−
1
7π
exp i
4
4
=
=−
i
;
4
i
.
4
(c) Clearly, the singularity at z = π is removable, so
res(f ; π) = 0.
Since
sin z
→1
z
for z → 0,
the singularity at z = 0 is a simple pole, so
res(f ; 0) = lim z · f (z) = lim
z→0
z→0
sin z
1
1
·
= .
z
π−z
π
Alternatively we consider z = 0 as a pole of at most order 2, so it follows by Rule I that
res(f ; 0) =
1
d
lim
1! z→ dz
sin z
π−z
= lim
z→0
cos z
sin z
+
π−z
(π − z)2
=
1
.
π
Analogously we can consider z = π as a “pole” of at most order 1. Then by Rule I,
res(f ; π) = lim
z→π
−
sin z
z2
= 0.
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13
Complex Funktions Examples c-6
Residues in finite singularities
Example 2.8 Find the residues of all singularities in C of
(a)
z ei z
,
(z − π)2
(b)
z3 + 5
,
(z 4 − 1) (z + 1)
(c)
ez
.
z3 − z
(a) The only singularity is a double pole at z = π, so if follows from Rule I that
res
z ei z
;π
(z − π)2
=
d
1
z ei z = lim ei z + i z ei z = −1 − i π.
lim
z→π
z→π
dz
1!
(b) The function
f (z) =
z3 + 5
(z 4 − 1) (z + 1)
has the three simple poles 1, i and −i, and the double pole −1. If we put
z3 + 5
z+1
and
B(z) = z 4 − 1,
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14
Complex Funktions Examples c-6
Residues in finite singularities
where the simple poles z0 = 1, i, −i, all satisfy z04 = 1, then
res (f ; z0 ) =
1 z0 z03 + 5
1 1 + 5z0
1
z0
A (z0 )
· 4·
=
= ·
= +
,
′
B (z0 )
4 z0 z0 + 1
4 1 + z0
4 1 + z0
hence
res(f ; 1)
=
3
1 1
+ = ;
4
4 2
res(f ; i)
=
i
3 + 2i
1
+
=
;
4
4 1+i
res(f ; −i)
=
1
i
3 − 2i
−
=
.
4 1−i
4
Finally, it follows for the double pole −1 by Rule I,
res(f ; −1)
=
=
=
d
z→−1 dz
lim
lim
z→−1
(z 2
z3 + 5
(z 2 + 1) (z − 1)
2z z 3 + 5
z3 + 5
3z 2
− 2
−
2
+ 1) (z − 1) (z 2 + 1) (z − 1) (z + 1) (z − 1)2
4
3
2(−1) · 4
3
1
9
−
− 2
=− −1− =− .
2
2 · (−2) 2 · (−2) 2 · (−2)
4
2
4
Check. The sum of the residues is
3 3 + 2i 3 − 2i 9
+
+
− = 0.
4
4
4
4
This agrees with the fact that the function has a zero of second order at ∞, so the residuum in ∞
(the additional term) is 0 in this case.
(c) The poles z = −1, 0, 1 are all simple. Therefore we get by Rule I,
res(f ; −1)
=
res(f ; 0)
=
res(f ; 1)
=
ez
z→−1
z→−1 z(z − 1)
ez
= −1,
lim z f (z) = lim 2
z→0
z→0 z − 1
ez
=
lim (z − 1)f (z) = lim
z→1
z→1 z(z + 1)
lim (z + 1)f (z) = lim
=
1
e−1
= ,
(−1)(−2)
2e
e
.
2
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15
Complex Funktions Examples c-6
Residues in finite singularities
Example 2.9 Find the residues at z = 0 of the following functions:
(a)
2z + 1
,
z (z 3 − 5)
ez
.
sin z
(b)
We have in both cases a simple pole atz = 0. As usual there are several possibilities of solutions, of
which we only choose one.
(a) It follows by Rule I,
2z + 1
;0
z (z 3 − 5)
res
= lim
z→0
2z + 1
1
=− .
5
z3 − 5
(b) In this case Rule II is the easiest one:
ez
;0
sin z
res
ez
= 1.
z→0 cos z
= lim
Example 2.10 Find the residuum at z = 1 of
1
,
zn − 1
n ∈ N.
Here z = 1 is a simple pole, so by rule II,
res
zn
1
;1
−1
= lim
z→1
1
1
= .
n−1
nz
n
Addition. Let z0 denote any one of the simple poles, i.e. z0n = 1. Then it follows by Rule II that
res
1
; z0
zn − 1
=
z0
z0
1
= .
·
n
n z0n−1 z0
♦
Example 2.11 Find the residues at all singularities in C of
(a)
1
,
2
(z − 1) (z + 2)
(b)
z 3 − 1 (z + 2)
(z 4
− 1)
2
,
(c) exp
1
z−1
.
(a) The poles at −2, −1 and 1 are all simple, hence by Rule I,
res
res
res
1
; −2
(z 2 − 1) (z + 2)
=
1
; −1
− 1) (z + 2)
=
1
;1
(z 2 − 1) (z + 2)
=
(z 2
lim
z→−2
1
1
= ,
z2 − 1
3
1
1
=− ,
z→−1 (z − 1)(z + 2)
2
lim
lim
z→1
1
1
= .
(z + 1)(z + 2)
6
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16
Complex Funktions Examples c-6
Residues in finite singularities
Remark 2.1 Here,
res
1
; −2 + res
(z 2 − 1) (z + 2)
1
; −1 + res
(z 2 − 1) (z + 2)
1
;1
(z 2 − 1) (z + 2)
= 0,
in agreement with the fact that we have a zero of order 3 at ∞, so the residuum here (the additional
term) is 0. ♦
(b) The poles are her z = 1, i, −1, −i, and z = 1 is a simple pole, while the other ones are double
poles. Hence by various applications of Rule I,
z3 − 1
· (z + 2)
z − 1 (z + 2)
3z 2 (z + 2)
9
z−1
=
lim
=
lim
=
,
lim (z − 1)
2
2
2
4
4
3
z→1
z→1
z→1
16
(z − 1)
(4z )
z −1
z−1
3
res(f ; 1)
=
z 3 − 1 (z + 2)
d
z→−1 dz
res(f ; −1) = lim
=
=
lim
z→−1
3z 2 (z+2)+z 3 −1
2
(z 2 +1) (z−1)2
d
z→i dz
=
=
lim
3
(z 2 +1) (z−1)2
z→i
(z+i)2 (z 2 −1)
(z+i)2 (z 2 −1)
−
2 z 3 −1 (z+2)
2
(z 2 +1) (z−1)3
2
−
2
2 z 3 −1 (z+2)
(z+i)2 (z 2 −1)
2
−
2 · 2z z 3 −1 (z+2)
(z+i)2 (z 2−)
3
−3(2 + i) − i − 1 2 · (−1 − i)(2 + i) 4i(−1 − i)(2 + i)
−
−
−4 · (−2)2
−8i(−2)2
−4(−2)3
7 + 4i −1 − 3i 2i(−1 − 3i)
7 + 4i −3 + i −6 + 2i
−2 + 7i
+
−
=
+
+
=
,
16
16i
16
16
16
16
16
d
z→−i dz
=
2 · 2z z 3 −1 (z+2)
z 3 −1 (z+2)
3z 2 (z+2)+z 3 −1
res(f ; −i) = lim
=
−
3 · 1 − 1 − 1 2 · 2 · (−1) · (−2) · 1 2 · (−2) · 1
1
4
2
5
−
− 2
=
−
−
=− ,
22 · (−2)2
23 · (−2)2
2 · (−2)3
16 16 16
16
res(f ; i) = lim
=
2
(z 2 + 1) (z − 1)2
lim
z→−i
z 3 −1 (z+2)
(z−i)2 (z 2 −1)
3z 2 (z+2)+z 3 −1
(z−i)2 (z 2 −1)
2
−
2
2 z 3 −1 (z+2)
(z−i)3 (z 2 −1)
2
−
2 · 2z z 3 −1 (z+2)
(z−i)2 (z 2 −1)
3
−2 − 7i
−3(2 − i) + i − 1 2(−1 + i)(2 − i) (−4i)(−1 + i)(2 − i)
−
−
= res(f ; i) =
.
2
2
3
−4 · (−2)
8i(−2)
−4 · (−2)
16
Note again that the sum of residues is 0.
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17
Complex Funktions Examples c-6
Residues in finite singularities
(c) The only singularity here is z = 1. It is essential, so we must expand into a Laurent series from
z0 = 1,
exp
1
z−1
=1+
1
+ ··· ,
z−1
, z = 1.
It follows that
res exp
1
z−1
;1
= a−1 = 1.
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18
Complex Funktions Examples c-6
Residues in finite singularities
Example 2.12 Prove that the functions
(a)
1
,
sin z
(b)
1
,
1 − ez
only have simple poles in C. Find these end their corresponding residues.
1
are the same as the zeros of sin z and of the same multiplicity. The function
sin z
sin z has the zeros {p π | p ∈ Z}, where
(a) The poles of
lim
z⊤ π
d
sin z = lim cos z = (−1)p = 0,
z→p π
dz
hence all poles are simple. Finally, it follows by Rule II that
res
1
; pπ
sin z
= lim
z⊤ π
1
= (−1)p ,
cos z
p ∈ Z.
1
are the same as the zeros of 1 − ez and of the same multiplicity. The zeros
1 − ez
are z = 2i p π, p ∈ Z, and since
(b) The poles of
d
(1 − ez ) = −ez = 0
dz
for every z ∈ C,
all poles are simple. Hence by Rule II,
res
1
; 2i pπ
1 − ez
=
lim
z→2i p π
1
= −1,
−ez
p ∈ Z.
Example 2.13 Find the residues at all singularities in C of
1
.
1 − cos z
1
has a (non-isolated) essential singularity at ∞, and otherwise only poles in C.
1 − cos z
The poles are determined by the equation 1 − cos z = 0, thus
The function
z
(1) 0 = 1 − cos z = 2 sin2 ,
2
the complete solution of which is z = 2pπ, p ∈ Z. It follows from (1) that the zeros are all of second
1
order, hence the poles z = 2pπ, p ∈ Z, of
are all of second order. We then have by Rule I
1 − cos z
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19
Complex Funktions Examples c-6
Residues in finite singularities
and l’Hospital’s Rule the following dreadful computation,
1
1
d (z − 2pπ)2
; 2pπ =
lim
1! z→2pπ dz
1 − cos z
1 − cos z
2(z−2pπ)(1−cos z)−(z− 2pπ)2 sin z
=
lim
z→2pπ
(1−cos z)2
res
=
=
lim
z→2pπ
1
2
2(z−2pπ) · 2 sin2
lim
z→2pπ
2 2·
z
2
z
2
− (z−2pπ)2 · 2 sin z2 cos z2
z 2
2
2 sin2
− 2pπ sin z2 − 2 ·
sin3
z
2
z
2
− 2pπ
2
cos z2
(w − pπ) sin w − (w − pπ)2 cos w
w→pπ
sin3 w
sin w+(w−pπ) cos w−2(w−pπ) cos w−(w−pπ)2 sin w
2 lim
w→pπ
3 sin2 w · cos w
2
sin w − (w − pπ) cos w + (w − pπ)2 sin w
lim
3 w→pπ
sin2 w · cos w
2
cos w−cos w+(w−pπ) sin w+2(w−pπ) sin w+(w−pπ)2 cos w
lim
3 w→pπ
2 sin w · cos2 w − sin3 w
3(w−pπ) sin w+(w−pπ)2 cos w
2
lim
3 w→pπ
2 sin w · cos2 w − sin3 w
3 sin w+3(w−pπ) cos w+2(w−pπ) cos w−(w−pπ)2 sin w
2
lim
3 w→pπ
2 cos3 w−4 sin2 w · cos w−3 sin2 w · cos w
0.
= 2 lim
=
=
=
=
=
=
Remark 2.2 Whenever one apparently has to go through some heavy computations like the previous
ones, one should check if there should not be another easier method. Here it would have been cheating
the reader first to bring the simple solution, so for pedagogical reasons we have first given the standard
solution.
An alternative method of solution is the following: First note that we have for every z ∈ C and
every p ∈ Z that
cos((z + 2pπ) − 2pπ) = cos(−(z + 2pπ) − 2pπ),
which is just another way of saying that the function 1 − cos z is an even function with respect to any
2pπ, p ∈ Z, so if we expand the function from some 2pπ, then it is again even. In a Laurent series
expansion of any even function all coefficients a2n+1 , n ∈ Z, of odd indices must be equal to 0. In
particular,
res
1
; 2pπ
1 − cos z
= a−1 = 0
for ethvert p ∈ Z.
♦
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20
Complex Funktions Examples c-6
Residues in finite singularities
Example 2.14 Find the residues at all singularities in C of
sinh z
.
sin2 z
Clearly, the poles are z = pπ, p ∈ Z, and z = 0 is a simple pole. Any other pole z = pπ, p ∈ Z \ {0}
is a double pole.
When we apply Rule I, we get
z · sinh z
= lim
z→0
z→0 sin2 z
res(f ; 0) = lim
z z+
z3
+ ···
3!
z3
z−
+ ···
3!
z2 1 +
2
= lim
z→0
z2
z2
+ ···
6
z2
+ ···
1−
6
2
= 1,
and
res
sinh z
; pπ
sin2 z
=
d
1
lim
1! z→pπ dz
(z − pπ)2 sinh z
sin2 z
z 2 sinh(z + pπ)
sin2 z
d
z→0 dz
= lim
= a1 ,
where we to ease matters have put
(2)
z 2 sinh(z + pπ)
= a0 + a1 z + · · · , |z| < π,
sin2 z
because z = 0 is a removable singularity, and the function has a Taylor expansion in the open disc of
centrum 0 and radius π.
The task is now to determine the coefficient a1 in the Taylor expansion. It is obvious that the usual
definition with a differentiation followed by taking a limit becomes very messy. Instead we multiply
by the denominator, so (2) becomes equivalent to
z 2 sinh(z + pπ) = (a0 + a1 z + · · · ) sin2 z = (a0 + a1 z + · · · ) ·
1
(1 − cos 2z),
2
hence after insertion of the series expansions,
z 2 {sinh pπ + cosh pπ · z + · · · } = (a0 + a1 z + · · · ) ·
= (a0 + a1 z + · · · ) z 2 −
1 4
z + ···
3
1
2
1−1+
1
1
(2z)2 − (2z)4 + · · ·
4!
2
,
which for z = 0 is reduced to
sinh pπ + cosh pπ · z + · · · = (a0 + a1 z + · · · ) 1 −
1
+ ···
3
= a 0 + a1 z + · · · .
When we identify the coefficients, we get
a0 = sinh pπ
and
a1 = cosh pπ,
so we conclude that
res
sinh z
; pπ
sin2 z
= a1 = cosh pπ,
res
sinh z
;0
sin2 z
= 1 = cosh(0 · π),
p ∈ Z \ {0},
p = 0,
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21
Complex Funktions Examples c-6
Residues in finite singularities
(cf. the above). Summing up we have in general,
res
sinh z
; pπ
sin2 z
= cosh pπ,
p ∈ Z.
Example 2.15 Find all Laurent series solutions in a disc with the centrum z 0 = 0 excluded of the
differential equation
z 4 + z 2 f ′ (z) + 2 z 3 + z f (z) = 1,
and find the value of the complex line integral
f (z) dz
|z|= 12
for everyone of these solutions.
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z 2 f ′ (z) + 2z f (z) = z 2 + 1
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22
Complex Funktions Examples c-6
Residues in finite singularities
When |z| < 1 this equation can be written
d
dz
+∞
z 2 f (z) =
1
(−1)n z 2n ,
=
1 + z2
n=0
hence by termwise integration in the open unit disc |z| < 1:
+∞
z 2 f (z) = C +
(−1)n 2n+1
z
= C + Arctan z,
2n + 1
n=0
C ∈ C arbitrary constant,
and the complete solution in the disc (without its centrum) is given by
+∞
f (z) =
(−1)n 2n−1
C
C
Arctan z
z
= 2+
+
,
2
2n + 1
z
z
z2
n=0
The circle z| =
|z|= 12
C ∈ C,
0 < |z| < 1.
1
lies in this set, so we conclude that
2
f (z) dz = 2πi · res(f ; 0) = 2πi a−1 = 2πi,
which holds for all of the solutions above.
Second method. The method of series. The coefficient z 4 + z 2 = z 2 z 2 + 1 is 0 for z = 0 or for
z = ±i, and the solution f (z) is analytic in its domain. Therefore, we get by inserting the Laurent
series
f (z) =
an z n ,
f ′ (z) =
n an z n−1 ,
into the differential equation that
z 4 + z 2 f ′ (z) + 2 z 3 + z f (z)
=
n an z n+3 +
n an z n+1 +
=
(n + 2)an z n+3 +
=
n an−2 z n+1 +
=
{n an−2 + (n + 2)an } z n+1 .
2an z n+3 +
2an z n+1
(n + 2)an z n+1
(n + 2)an z n+1
This expression is the identity theorem equal to 1, if −a−3 + a−1 = 1 and the following recursion
formula holds,
n an−2 + (n + 2)an = 0,
for n ∈ Z \ {−1}.
If n = 0, then a0 = 0 and a−2 is an indeterminate. Then it follows by recursion that a2n = 0 for
n ∈ N0 .
If n = −2, then a−4 = 0 and a−2 is an indeterminate. It follows by recursion that a−2n = 0 for
n ∈ N \ {1}.
It only remains to find the coefficients of odd indices, where we have already proved that
−a−3 + a−1 = 1.
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23
Complex Funktions Examples c-6
Residues in finite singularities
We have for the odd indices the recursion formulæ
a2n−1 = −
2n − 1
a2n−3 ,
2n + 3
n ∈ N,
and
a−2n−3 = −
−2n + 1
a−2n−1 ,
−2n − 1
n ∈ N.
Hence by recursion for the positive, odd indices,
a2n−1 = −
2n − 1
2n − 1 2n − 3
3 1
(−1)n
· · · · · a−1 =
a2n−3 = · · · = (−1)n ·
·
,
2n + 1
2n + 1 2n − 1
2n + 1
5 3
where the corresponding series is convergent for 0 < |z| < 1. This series is determined by the
coefficient a−1 .
The analogous coefficients corresponding to the negative odd indices ≤ 3 have a similar structure,
1
< 1, i.e. the set given by |z| > 1. This
corresponding to the domain of convergence given by
z
series is determined by the coefficient a−3 .
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