Complex Analysis
Topic: Singularities
MA201 Mathematics III
Department of Mathematics
IIT Guwahati
August 2015
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Zeroes of Analytic Functions
A point z0 ∈ C is a zero of order (or multiplicity) m (m ≥ 1) of f if
f (z0 ) = f 0 (z0 ) = · · · = f (m−1) (z0 ) = 0, and f (m) (z0 ) 6= 0.
A zero of order 1 is called a simple zero.
Examples.
For f (z) = z(z − 1)5 : z0 = 0 is a simple zero and z0 = 1 is a zero
of order 5.
z0 = nπ, n ∈ Z are simple zeroes of f (z) = sin(z).
f (z) = (z 2 − πz) sin(z) has zeroes of order two at z0 = 0 and π.
f (z) = ez has no zeroes in C.
Every polynomial of order n ≥ 1 has zeroes in C with their total
multiplicity n.
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If z0 is a zero of a polynomial P (z) of multiplicity m, then
P (z) = (z − z0 )m Q(z) where Q(z) is a polynomial. In general, we have
the following:
Theorem
Suppose f is analytic in a domain D, and z0 ∈ D. Then z0 is a zero of
order m for f if and only if f (z) = (z − z0 )m g(z) for some function g
which is analytic in a neighborhood of z0 and g(z0 ) 6= 0.
Proof. ⇒) Follows from the Taylor series
!
∞
X
f (n) (z0 )
f (z) =
(z − z0 )n .
n!
n=0
P∞
n
n=0 bn (z − z0 ) , where b0 = g(z0 ) 6= 0. Then,
f (z) = b0 (z − z0 )m + b0 (z − z0 )m + · · · must be the Taylor series of f (z).
So, f (z0 ) = f 0 (z0 ) = · · · = f (m−1) (z0 ) = 0 and f (m) (z0 ) = (m!)b0 6= 0.
⇐) g(z) =
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Theorem
For a function f , analytic on a domain D, the following are equivalent.
1
There is z0 ∈ D such that f (n) (z0 ) = 0 for n ≥ 0.
2
f ≡ 0 in D.
3
The set S = {z ∈ C : f (z) = 0} has a limit point in D.
Proof. (1) ⇒ Taylor series about z0 is the zero series
⇒ f (z0 ) ≡ 0 in an open disc centered at z0 .
⇒ given z ∈ D keep shifting the disc towards z and get f (z) = 0 ⇒ (2)
(2) ⇒ (3) Trivial.
(3) ⇒ (1) Suppose z0 is a limit point of S. We claim that f (n) (z0 ) = 0 for
all n ≥ 0. Suppose not. Let m be least such that f (m) (z0 ) 6= 0. Then
f (z) = (z − z0 )m g(z) where g is analytic and g(z0 ) 6= 0. There is r > 0
such that g(z) 6= 0 in Br (z0 ), because g is continuous. Thus, f (z) 6= 0
for 0 < |z − z0 | < r, a contradiction.
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Corollaries
For a non-constant function f on a domain D
A zero of infinite order is not possible.
Set of zeroes of f cannot have a limit point in D, i.e., each zero of f
is isolated.
If two functions f and g, analytic in a domain D, are such that the
set {z ∈ D : f (z) = g(z)} has a limit point, then f (z) = g(z) for all
z ∈ D.
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A point z0 ∈ C is a singular point or a singularity of a function f , if
1
f is not analytic at z0 , and
2
every neighborhood of z0 contains a point at which f is analytic.
1
has a singular point z0 = 0.
z
f (z) = cot(z) has singular points z = nπ, n ∈ Z.
Example. f (z) =
Points on the negative real axis are singular points for
f (z) = Log (z).
A singular point z0 ∈ C is an isolated singular point of f , if there is a
deleted neighborhood 0 < |z − z0 | < r of z0 throughout which f is
analytic. Otherwise, z0 is called a non-isolated singular point.
1 1/z sin(z)
, e ,
.
z
z
z0 = nπ are isolated singular points for cot(z).
0 and the points on the negative real axis are non-isolated
singular points for Log (z).
Example. z0 = 0 is an isolated singular point for
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Classification of isolated singular points
Let z0 be a singular point of f . Then
1
z0 is called a removable singular point, if lim f (z) exists and
z→z0
equals a complex number a0 .
Example: z0 = 0 is a removable singular point of
2
sin z
ez − 1
and
.
z
z
z0 is called a pole, if lim f (z) = ∞.
z→z0
1 1
,
and cot z.
z z2
z0 is called an essential singular point, lim f (z) does not exist.
z→z0
1
1/z
Example: z0 = 0 is an essential singular point of e and sin
.
z
Example: z0 = 0 is a pole for
3
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Order of a pole
Note: A point z0 is a pole of f if and only if z0 is a zero of 1/f .
Definition
A pole z0 of f is said to be of order m ≥ 1 if z0 is a zero of order m of
the function 1/f .
A pole of order 1 is called a simple pole.
Examples:
cot z has simple poles at the points z0 = nπ, n ∈ Z.
z0 = 1 is a pole of order 2 for the function f (z) =
1
.
(z − 1)2
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Singular points and Laurent series
Theorem
Suppose f has an isolated singular point at z0 . Let
f (z) =
∞
X
an (z − z0 )n
n=−∞
be the Laurent series of f about z0 , 0 ≤ |z − z0 | < r for r > 0. Then
1
z0 is a removable singularity if and only if a−n = 0 for all n ∈ N.
2
z0 is a pole of order m if and only if a−m 6= 0 and a−n = 0 for all
n > m.
3
z0 is an essential singularity if and only if a−n 6= 0 for infinitely
many n ∈ N.
Note: If f has a removable singularity at z0 , then defining
f (z0 ) = lim f (z) = a0 , f becomes analytic at z0 .
z→z0
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Examples
ez − 1
z
z2 z3
=1+ +
+
+ · · · for 0 < z < ∞.
z
2!
3!
4!
So, z0 = 0 is a removable singularity of f .
f (z) =
Note: Defining f (0) = 1 you get f analytic at z0 = 0.
∞
f (z) =
X
1
−1
=
−
(z − 1)n for 0 < |z − 1| < 1.
(z − 1)(z − 2)
z−1
n=0
So, z0 = 1 is a simple pole of f . Similarly, z0 = 2 is a simple pole
of f .
n
∞
1
1 X
1
1
2
2
f (z) = z exp z = z + z + +
for |z| > 0.
2!
(n + 2)! z
n=1
So, z0 = 0 is an essential singularity of f .
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Properties of Removable Singularity
Theorem
Suppose that f is analytic in 0 < |z − z0 | < r for some r > 0, and z0 is a singular point
of f . Then the following are equivalent:
1
2
z0 is a removable singularity of f .
∞
X
f (z) =
an (z − z0 )n for 0 < |z − z0 | < r.
n=0
3
4
There is an analytic function g in |z − z0 | < r such that g(z) = f (z) for
0 < |z − z0 | < r.
lim f (z) exists and is finite.
lim (z − z0 )f (z) = 0.
z→z0
5
z→z0
6
f is bounded in a deleted neighborhood of z0 .
∞
X
a−n
+
an (z − z0 )n for 0 < |z − z0 | < r.
(z − z0 )n n=0
n=1
must be zero for all n ∈ N.
Look at the Laurent series f (z) =
If (5) or (6) holds, then a−n
∞
X
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Properties of a Pole
Theorem
Suppose that f is analytic in 0 < |z − z0 | < r for some r > 0. Then the following are
equivalent:
1
2
3
f has a pole of order m at z0 .
1
has a zero of order m at z0 .
f
∞
X
f (z) =
ak (z − z0 )k for 0 < |z − z0 | < r and a−m 6= 0.
k=−m
g(z)
, where g is analytic at z0 and g(z0 ) 6= 0.
(z − z0 )m
4
f (z) =
5
The function (z − z0 )m f (z) has a removable singularity at z0 .
6
limz→z0 |f (z)| = ∞.
7

 0,
a−m ,
lim (z − z0 ) f (z) =
z→z0

∞,
k
if k > m,
if k = m,
if k < m.
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Properties of Essential Singularities
Theorem
Suppose that f is analytic in 0 < |z − z0 | < r for some r > 0. Then the following are
equivalent:
1
f has essential singularity at z0 , i.e., lim does not exist.
z→z0
2
f (z) =
∞
X
ak (z − z0 )k for 0 < |z − z0 | < r and a−n 6= 0 for infinitely many
k=−∞
n ∈ N.
3
f is neither bounded nor approaches ∞ as z → z0 .
Theorem
Picard’s Great Theorem: Suppose has an essential singularity at z0 . Then in each
deleted neighborhood N (z0 ) of z0 , f assumes every complex number, with possibly
one exception, an infinite number of times, i.e., the image of 0 < |z − z0 | < r under f
is an infinite number of copies of C \ {atmost one point}.
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Singularity at Point at Infinity (∞)
Definition
Suppose that
f is analytic in R < |z| < ∞ for some R ≥ 0. Define g by
1
g(z) = f z for 0 < |z| < 1/R. Then, f is said to have a removable
singularity / pole of order m / essential singularity at ∞ if g has a
removable singularity / pole of order m / essential singularity at 0,
respectively.
Examples. (1) If P (z) is a non-constant polynomial, then the function
1
f given by f (z) =
has a removable singularity at z = ∞.
P (z)
(2) Any non-constant polynomial P (z) has a pole at z = ∞.
(3) ez , sin(z), cos(z), sinh(z) have an essential singularity at z = ∞.
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Result
An entire function has removable singularity at ∞ if and only if it is a
constant function.
[Hint. For only if part, show that f is bounded.]
Result
An entire function has a pole of order m at ∞ if and only if it is a
polynomial of degree m.
Exercise. Analyze the nature of the point z = ∞ for the rational
P (z)
function R(z) =
, where P (z) and Q(z) are polynomials of
Q(z)
degrees m and n, respectively.
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