11.1 The Laurent expansion of an analytic function
Theorem 88 Given a series c0 + c1
(z a) + c2
(z a)2 + ::: + cn
(z a)n + ::::;
and
l = lim
n!1
pn
jcnj:
Then the following are true:
i). l = 0 means the series is absolutely convergent for all z on the extended complex plane except for z = a;
ii). 0 < l < 1 means the series is absolutely convergent for all z outside the circle jz aj = l and divergent for all z inside the same circle; and
iii). l = 1 means the series is divergent for all nite z on C:
pf. Use the change of variable = 1= (z a) : Then let's alter the series into
c0 + c1 + c2 2 + ::: + cn n + ::::;
with radius of convergence R = 1
l = lim
n!1
pn
jcnj
1
;
Complex Analysis 153 and carries the points z = a and z = 1 into = 1 and = 0;
respectively.
i). By theorem, the altered series is absolutely convergent for j j < 1l and is divergent for j j > 1l; i. e. the original series converges absolutely for jz aj > l and is divergent for jz aj < l:
ii). When l = 0; R = 1 : the altered series converges for all nite ; i. e. the original series converges for all z except for z = a:
iii). Whn l = 1; R = 0 : the altered series converges only at = 0; i. e. the original series diverges for all nite z except for z = 1:
Remark 19 The transformation = 1= (z a) carries every bounded closed domain D outside jz aj = l into a closed bounded domain D0 inside j j = 1l:
Figure.
Since c0+ c1 + c2 2+ ::: + cn n+ :::: is uniformly convergent on D0; we nd that the series
c0 + c1
(z a) + c2
(z a)2 + ::: + cn
(z a)n + ::::
Complex Analysis 154 is analytic outside the circle jz aj = l: When l = 0; the series in z is analytic on the whole z plane except for z = a:
Def.
X1 n= 1
cn (z a)n ; 8cn 2 C;
is called a ”Laurent series.” Note that a Laurent series is the sum of two series
X1 n=0
cn(z a)n and
X1 m=1
c m (z a) m ;
where the prior series is called the ”regular part” of Laurent series, and the later series is called the ”principle part” of Laurent series.
Remark 20 The regular part of the Laurent series stated above is with radius of convergence
R = lim
n!1
pn
jcnj 1;
and the principle part is with radius of convergence r = lim
m!1
pm
jc mj:
Assume that r < R; then Laurent series is absolutely convergent and analytic on the annulus r < jz aj < R:
Complex Analysis 155
pf. For the given f (z) ; we can evaluate the integral as 1
(z a)n k+1 also converges uniformly on C. Hence we may interchange the order of integration and sum of the equation:
1
Complex Analysis 156
By the above discussions, we realize that the integral X1 with coef cients cn such that
cn = 1
Complex Analysis 157 Theorem 90 If f (z) is analytic on the annulus
Aa = fzj r < jz aj < Rg ; then f (z) has a Laurent expansion at z = a:
pf. 8 2 Aa; consider 3 circles: C (enclosing ); ; and 9 ; r0; R0 and l such that
r < r0 < j aj = < R0 < R and 0 < l < R0 r0 2 ; where C = fzj jz j = lg ; = fzj jz aj = r0g and
= fzj jz aj = R0g : These circles are all considered in the counterclockwise direction. Then we nd
Figure.
1 2 i
I
C
f (z)
(z a)n+1dz = 1 2 i
I f (z)
(z a)n+1dz
I f (z)
(z a)n+1dz ; 8n 2 Z:
Complex Analysis 158 By Cauchy's integral formula, we nd
f ( ) = 1
(z a)n+1 is uniformly convergent on : ii). 8z 2 ;
Complex Analysis 159 P1
n=1
f (z)( a) n
(z a) n+1 is uniformly convergent on : Combine i). and ii)., we obtain that
f ( ) =
Complex Analysis 160 iii). 1 < jz 1j < 1 :
f (z) = 1
z 1
1
1 + (z 1) = 1 (z 1)2
1 1 + z 11
= 1
(z 1)2
" 1 X
n=0
( 1)n 1
z 1
n#
= 1
(z 1)2 + 1 (z 1)3
1
(z 1)4 + :::::: ]
Complex Analysis 161 11.2 Isolated singular point
Def. We call z0 a ”singular point” of f (z) if f (z) is not analytic at z0: Also z0 is called an ”isolated singular point” of f (z) if f is analytic in some neighborhood of z0 except for z0 itself.
Def. If z0 is an isolated singular point of f (z) and f (z) has a Laurent expansion on some deleted neighborhood of z0; i. e.
f (z) =
X1 n= 1
cn (z z0)n ; 0 < jz z0j < R:
Then i). The series above contains no term of negative powers of (z z0) ; in this case, z0 is called a ”removable singular point” of f (z):
ii). The series above contains only a nite number of negative powers of (z z0) ; in this case, z0 is called a ”pole” of f (z):
iii). The series above contains in nitely many negative powers of (z z0) ; then z0 is called an ”essential singular point”
of f (z):
Note: Usually we like to state the term ”singular point” as
”singularity” in short.
Complex Analysis 162 11.2.1 Removable singularity
In case z0 is a removable singularity of f (z) ; then f (z) =
X1 n=0
cn(z z0)n ; 0 < jz z0j < R:
Notice that the series at the right hand side is analytic on jz z0j < R: Therefore we re-di ne f (z) such that
f (z) =
X1 n=0
cn (z z0)n ; which is analytic on jz z0j < R; where
f (z0) lim
z!z0 f (z) = c0: Ex. For the function
f (z) = sin z z ;
z = 0 is a removable singularity. Since the Taylor series for sin z is
sin z = z z3
3! + z5 5!
z7
7! + :::::; jzj < 1;
we have
sin z
z = 1 z2
3! + z4 5!
z6
7! + ::::; 0 < jzj < 1:
Complex Analysis 163 Re-de ne f (0) = limz!0 f (z) = 1; then
f (z) = 1 z2
3! + z4 5!
z6
7! + ::::; jzj < 1:
11.2.2 Pole
If z0 is a pole of f (z) ; the Laurent series expansion (in powers of z z0) of f (z) contains only nitely many negative powers of (z z0) : Let m be the highest power of z z1
0: Then f (z) =
X1 n=0
cn (z z0)n + c 1
z z0 + c 2
(z z0)2 + ::: + c m
(z z0)m;
where 0 < jz z0j < R for some R and c m 6= 0: We then call the point z0 the pole of order m of f (z) : When m = 1; z0 is a singular point; when m > 1; z0 is also called a multiple pole.
Note: If z0 is a pole of order m; we can write f (z) as (z z0)m f (z) =
X1 n=0
cn(z z0)m+n + c 1(z z0)m 1 + ::: + c m+1 (z z0) + c m
on the annulus 0 < jz z0j < R: In this case z0 is a removable singularity of (z z0)m f (z) and
zlim!z0 (z z0)m f (z) = c m 6= 0:
Complex Analysis 164 After observing that, we nd
zlim!z0 f (z) = lim
z!z0 (z z0)m f (z) 1
(z z0)m = 1:
i. e. f (z) is unbounded on any deleted neighborhood of z0 when z0 is a pole of f (z):
Ex. Explain why z = 0 is a pole of the function g (z) = cos z
z :
Sol. The Taylor series expansion for cos z is cos z = 1 z2
11.2.3 Realtion between zeros & poles
Theorem 91 Let z0 be a zero of order m of a function f (z) ana-lytic at z0: Then 1=f (z) is analytic on a deleted neighborhood of z0 with a pole of order m at z0:
Complex Analysis 165 pf. According to the description of f; we have
f (z) =
X1 n=m
cn(z z0)n = (z z0)m
" 1 X
n=0
cn+m (z z0)n
#
on some neighborhood of z0; i. e. jz z0j < R for some R; and note that cm 6= 0: Let
' (z) = cm + cm+1(z z0) + cm+2(z z0)2 + :::::;
jz z0j < R for some R: Then we nd that ' (z0) = cm 6= 0 and ' is continuous at z0: So 9 some neighborhood of z0; say jz z0j < R0 R; such that ' (z) 6= 0 in jz z0j < R0: Thus set
(z) = 1 ' (z);
which is analytic on jz z0j < R0; and obviously, we may write (z) in terms of a Taylor expansion at z0; i. e.
(z) = (z0) + 0 (z0) (z z0) + 00 (z0) (z z0)2
2! + :::::
Finally, we obtain 1
f (z) = (z0)
(z z0)m +
0 (z0)
(z z0)m 1 + :::::
on 0 < jz z0j < R0: This shows that 1=f (z) has a pole of order m at z0:
Theorem 92 Let z0 be a pole of order m of a function f (z)
an-Complex Analysis 166 alytic in a deleted neighborhood of z0: Then 1=f (z) is analytic at z0 provided that 1=f (z0) = 0 with a zero of order m at z0:
pf. f (z) has a Laurent series expansion f (z) =
X1 n=0
cn (z z0)n + c 1
z z0 + c 2
(z z0)2 + ::: + c m
(z z0)m;
where 0 < jz z0j < R for some R: Then again, (z z0)m f (z) =
X1 n=0
cn(z z0)n+m + c 1(z z0)m 1 + :::: + c m
with 0 < jz z0j < R: Let ' (z) =
X1 n=0
cn (z z0)n+m + c 1(z z0)m 1 +c 2 (z z0)m 2 + :::: + c m; which is analytic on jz z0j < R: So it is obvious that (z z0)m f (z) = ' (z) on the annulus 0 < jz z0j < R:
Since ' (z0) = c m 6= 0 and ' is continuous at z0; we may nd ' (z) 6= 0 on jz z0j < R0 R for some R0: Then let
(z) = 1 ' (z);
Complex Analysis 167 which is analytic on jz z0j < R0: Consequently, (z) have a Taylor expansion
(z) = (z0) + 0 (z0) (z z0) + 00 (z0) (z z0)2
2! + :::::
for jz z0j < R0: Finally, look at 1
f (z) = (z z0)m
' (z) = (z z0)m (z)
= (z z0)m (z0) + 0 (z0) (z z0) + 00 (z0) (z z0)2
2! + :::::
#
for 0 < jz z0j < R0: Note that (z0) = 1
' (z0) = 1
c m 6= 0:
Then we see that z0 is a zero of order m of 1=f (z) :
Corollary 93 Let f (z) be non-vanishing in a deleted neighbor-hood of z0 and have an essential singularity at z0: Then 1=f (z) also has an essential singularity at z0:
pf. Suppose that z0 is not an essential singularity of 1=f (z);
i. e. z0 is either a pole or a removable singularity of 1=f (z): Let's consider the following:
i). z0 is a pole of order m:
Complex Analysis 168 By previous theorem, f (z) has a zero of order m at z0; provided that f (z0) = 0: Then z0 is a removable singularity of f (z) ; (! )
ii). z0 is a removable singularity of 1=f (z) :
Let ' (z) = 1=f (z) : Then limz!z0 ' (z) exists, i. e.
zlim!z0 ' (z) = 0 _ lim
z!z0 ' (z) 6= 0:
a). In case limz!z0 ' (z) = 0; de ne ' (z0) = 0; then z0 is a zero of ' (z) : By theorem 91, we nd that z0 is a pole of 1=' (z) = f (z) ; (! ) :
b). In case limz!z0 ' (z) 6= 0; let limz!z0 ' (z) = c0 6= 0:
De ne ' (z0) = c0; then ' is analytic at z0 and ' (z) 6= 0 in some neighborhood of z0: This implies that 1=' (z) is analytic at z0; i.
e. f (z) is analytic at z0;
f (z0) = 1
' (z0) = 1 c0:
Then z0 is a removable singularity of f (z); (! )
Theorem 94 (Casorati-Weierstrass theorem) Let z0 be an essen-tial singularity of f (z) and A be any complex number ( nite or in nite.) Then there exists a sequence f kg converging to z0 such that
klim!1f ( k) = A:
Complex Analysis 169 pf. i). If A = 1 :
Suppose that 9M such that jf (z)j M in some deleted neighborhood of z0; i. e. jf (z)j M and f (z) is analytic on 0 < jz z0j < R; and
f (z) =
X1 n= 1
cn (z z0)n; where
cn = 1 2 i
I
C
f (z)
(z z0)n+1dz; n 2 Z;
C = fzj jz z0j = ; 0 < < R:g By Cauchy's inequality, jcnj M
n:
When n < 0; let ! 0 we must have cn = 0: Therefore z0 is a removable singularity of f (z); (! )
Thus f (z) cannot be bounded in any deleted neighborhood of z0; which means 8k; 9 k with
0 < j k z0j < 1
k 3 jf ( k)j > k;
or lim
k!z0 f ( k) = 1 as k ! 1;
and limk!1 f ( k) = 1:
ii). If A is any nite complex number:
If any deleted neighborhood of z0 contains a point z such
Complex Analysis 170 that f (z) = A; it needs no proof. So let's assume that there is no z in any deleted neighborhood K of z0: Let
' (z) = 1
f (z) A; 8z 2 K:
Since z0 is an essential singularity of f (z) and f (z) A; z0 is also an essential singularity of ' (z) : Therefore there is a sequence f kg such that k ! z0 and limk!1 '( k) = 1; or
klim!1[f ( k) A] = 1:
Ex. Discuss the essential singularity of f (z) = exp (1=z) ; and nd the sequence guaranteed by Casorati-Weierstrass theorem with respect to various values of A:
Sol. According to the de nition, f (z) = ez = 1 + z + z2
2! + z3
3! + ::::; 8z:
Then when z 6= 0;
exp 1
z = 1 + 1
z + 1
2!z2 + 1
3!z3 + ::::; 0 < jzj < 1:
So z = 0 is the essential singularity of exp(1=z): Let's consider the following cases:
i). A = 1 :
Complex Analysis 171 Let k = k1 ! 0: Then
f ( k) = exp 1
k
= exp k ! 1 as k ! 1: ] ii). A = 0 :
Let k = k1 ! 0: Then f ( k) = exp 1
k
= exp( k) ! 0 as k ! 1: ] iii). A 6= 0 ^ A 6= 1 :
For A = exp (1=z) ; we have ln A = ln exp 1
z = 1
z 2k i , z = 1
ln A: Choose
k = 1
fln Ag0 + 2k i ! 0 as k ! 1;
and then
klim!1f ( k) = lim
k!1exp [fln Ag0 + 2k i] = A: ]
Theorem 95 If f (z) is univalent in a domain G; then f0(z) 6= 0
Complex Analysis 172 in G:
pf. Suppose that f0 (z0) = 0 for some z0 2 G: Since f is analytic at z0; we must have
f (z) = c0 + ck (z z0)k + ck+1(z z0)k+1 + ::::;
where k 2 in some neighborhood of z0: Set
(z) = ck + ck+1 (z z0) + ck+2(z z0)2 + :::::
for some r1 such that (z) 6= 0 over jz z0j r1:
Now, since f0 (z) is analytic at z0 and that z0 is a zero of f0(z); we can de nitely nd some r2 such that f0 (z) 6= 0 over 0 < jz z0j r2; for the zero of an analytic function must be isolated. Let's choose
r = minfr1; r2g ; then we must have
(z) 6= 0 over jz z0j r as well as
f0 (z) 6= 0 over 0 < jz z0j r:
Hence there is some
= min
jz z0j=r ck (z z0)k + ck+1 (z z0)k+1 + :::: > 0;
for (z z0)k (z) 6= 0 on jz z0j r:
Complex Analysis 173 Next, let a 6= 0 be any real number such that j aj < : According to Rouché's theorem,
f (z) (c0 + a) = a + ck (z z0)k + ck+1 (z z0)k+1 + ::::
and the function
(z z0)k (z) = ck (z z0)k + ck+1(z z0)k+1 + ::::
have the same number of zeros inside the circle jz z0j = r; i. e.
i). f (z) (c0 + a) has exactly k zeros inside jz z0j = r;
ii). [f (z) (c0 + a)]0 = f0 (z) 6= 0 over 0 < jz z0j r;
and
iii). f (z0) (c0 + a) = c0 c0 a = a 6= 0:
Thus each zero of f (z0) (c0 + a) inside the circle
jz z0j = r is simple; i. e. 9 distinct k points z1; z2; :::; zk inside jz z0j = r such that
f (zj) (c0 + a) = 0; or f (zj) = c0 + a;
for j = 1; 2; 3; :::; k; k 2: This contradict to the fact that f is 1 1:
Complex Analysis 174 11.3 Residues
Def. Let z0 be an isolated singularity of a function f (z) : Then by the ”residue of f (z) at z0;” denoted by
Resz=z0 f (z) ;
is meant the coef cient c 1 in the Laurent expansion f (z) =
X1 n= 1
cn (z z0)n:
Theorem 96 (Residue theorem) f (z) is analytic inside and on a piecewise smooth Jordan curve C, except for isolated singularities z1; z2; ... ; zN lying inside C. Then
I
C
f (z) dz = 2 i
XN k=1
Resz=zk f (z) :
pf. Let 1; 2;...., N be circles centered at the points z1; z2; ... ; zN; respectively, which are so small that they all lie inside C and does not intersect each other. Then
I
C
f (z) dz =
XN k=1
I
k
f (z) dz:
Complex Analysis 175 Suppose that the Laurent expansion of f (z) at zk is
f (z) =
X1 n= 1
ckn(z zk)n
on 0 < jz zkj < k for some k containing k: Since the series is uniformly convergent on k for all k; we nd that
I
by Cauchy's integral theorem.
Ex.
Complex Analysis 176 and so
= 2 i
(n 1)!ez
z=1
= 2 ie (n 1)!: Or,
ez = ez 1 e = e
"
1 + (z 1) + (z 1)2
2! + ::::
#
and ez
(z 1)n = e 1
(z 1)n + 1
(z 1)n 1
+ 1
2! (z 1)n 2 + ::: + 1
(n 1)! (z 1) + ::: ; from which we may see the same result.
Derivation: To calculate residues at poles without making explicit use of a Laurent series, consider the following:
i). Let z0 be a simple pole of f (z) : Then f (z) = c 1
z z0 + c0 + c1 (z z0) + ::::
on 0 < jz z0j < r for some r: Multiplying both sides by (z z0) ; we nd
(z z0) f (z) = c 1 + c0 (z z0) + c1(z z0)2 + ::::;
and
zlim!z0 (z z0) f (z) = lim
z!z0 [c 1 + c0(z z0)
Complex Analysis 177 +c1 (z z0)2 + ::::i
= c 1: ]
ii). Let z0 be a pole of order m of f (z) : Then f (z) = c m
(z z0)m + c m+1
(z z0)m 1 + :::: + c 1
z z0 +
X1 n=0
cn (z z0)n
on 0 < jz z0j < r0 for some r0: Multiplying both sides by (z z0)m ; we nd
(z z0)m f (z) = c m + c m+1 (z z0) + :::: + c 1(z z0)m 1 +
X1 n=0
cn (z z0)n+m ; and then differentiate both sides m 1 times,
dm 1
dzm 1 [(z z0)m f (z)] = (m 1)!c 1 +
X1 n=m 1
cnn (n 1) [n (m 1) + 1] (z z0)n+m (m 1) : So we obtain the equation
zlim!z0
dm 1
dzm 1 [(z z0)m f (z)] = (m 1)!c 1:
Complex Analysis 178 downstairs sin z has z = 0 as a zero of order 1: Therefore,
Resz=0
by using change of variable in complex numbers.
Complex Analysis 179
Then we may re-write cos x = z + 1z
Complex Analysis 180 where 0 < p < 1:
Complex Analysis 181