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This document contains detailed and exam-oriented notes on Singularities of Analytic Functions, prepared for undergraduate, postgraduate, and competitive mathematics examinations. Subject / Course: Complex Analysis – Singularities Level: BSc / BSc (Honours) / MSc Mathematics CSIR-NET / GATE / JAM / NBHM / TIFR Syllabus Coverage: As per UGC-NET / CSIR-NET / GATE and standard university syllabus Prepared by: MSc Mathematics (NIT) Prepared using standard textbooks and classroom notes
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A singularity (or singular point) of a complex function f (z) is a point z 0 in the complex plane where f (z) is not analytic, but it is analytic in some neighborhood around z 0 (except possibly at z 0 ).
Actual Meaning: It is a point where the function “misbehaves” and cannot be expressed as a convergent Taylor series.
How to Check:
0, etc.)
Examples:
Trap: Do not assume every singularity is a pole; check carefully using limits or Laurent series.
Negation: If f (z) is analytic at z 0 , it is not a singularity.
Definition: A singularity z 0 is removable if limz→z 0 f (z) exists and is finite.
Laurent Series: No negative powers.
Geometrical Meaning: Function has a “hole” at z 0 which can be patched.
Examples:
Definition: z 0 is a pole of order m if (z − z 0 )mf (z) is analytic and non-zero at z 0.
Formula: f (z) = (^) (z−g(zz 0 ))m , g(z 0 ) ̸= 0
Geometrical Meaning: Function blows up to infinity at z 0.
Examples:
Definition: z 0 is essential if it is neither removable nor a pole.
Laurent Series: Infinite number of negative powers.
Geometrical Meaning: Function behaves wildly; dense in complex plane (Great Picard Theorem).
Examples:
f (z) =
n=−∞
an(z − z 0 )n^ +
n=
an(z − z 0 )n
Remarks (deep):
Key fact (one-liner): For an isolated singularity z 0 ,
f (z) =
n=−∞
an(z − z 0 )n^ (convergent in some 0 < |z − z 0 | < R).
Classification is read directly from the coefficients an:
Residue: Res(f ; z 0 ) = a− 1. Knowledge of the principal part gives residues directly — crucial in contour integration.
Construction remark: If f has a pole of order m at z 0 , there exists analytic g with g(z 0 ) ̸= 0 such that f (z) = g(z)/(z − z 0 )m. Expand g in Taylor series and divide term- by-term to obtain Laurent form.
Example F.1.1.
f (z) = e
z (^) − 1 − z − z^2 2 z^3
at z = 0.
Goal: Show 0 is removable and find the analytic extension.
Solution. Use the Taylor series ez^ =
n=
zn n!
. Then
ez^ − 1 − z − z 22 =
n=
zn n!.
Divide by z^3 :
f (z) =
n=
zn−^3 n!
m=
zm (m + 3)!
This is a Taylor series (no negative powers) valid for all z. Hence z = 0 is removable and the analytic extension is
f˜ (z) =
m=
zm (m + 3)!,^ f˜ (0) =^1 3!.
Key points: The use of full Taylor expansion avoids mistaken truncation; limit exists
finite: lim z→ 0 f (z) =^1 6
Example F.1.2.
g(z) = sin^ z^ −^ z^ +^
z^3 6 z^5
at z = 0.
Solution. Use sin z =
n=0(−1)n^
z^2 n+ (2n + 1)!. Subtract^ z^ −^
z^3 6 to get series starting at^ z
Precisely,
sin z − z + z 63 =
n=
(−1)n^ z
2 n+ (2n + 1)!.
Divide by z^5 :
g(z) =
n=
(−1)n^ z
2 n− 4 (2n + 1)!,
which is analytic at 0. Hence removable; g(0) = coefficient of z^0 term which occurs at 2 n − 4 = 0 ⇒ n = 2: g(0) = (−1)^2 5!^1 = 1201.
Example F.1.3. (Hard rational-trig combination)
h(z) = tan^ z^ −^ z^ −^
z^3 3 z^5
at z = 0.
Solution sketch. Expand tan z = z + z 33 + 215 z^5 + O(z^7 ). Substituting gives numerator 2 z^5 15 +^ O(z
(^7) ). Divide by z (^5) to get h(z) = 2 15 +^ O(z
(^2) ). Thus removable with h(0) = 2/15.
□
Example F.2.1. (Pole of higher order via factorization)
f (z) = (^) (zcos − πz) 3 at z 0 = π.
Solution. Expand cos z about π: write z = π + w. Then
cos(π + w) = − cos w = −
1 − w
2 2 +^
w^4 24 − · · ·
So
f (z) =
1 − w 22 + w 244 − · · ·
w^3 =^ −^
w^3 +
w −^
w 24 +^ · · ·^. Hence z = π is a pole of order 3. Principal part is −w−^3 + 12 w−^1. Residue = a− 1 = 12.
Example F.2.2. (Rational function with cancellation — find order)
g(z) = e
z (z − 1)^4 (z − 2)
Goal: classify at z = 1 and z = 2.
Solution. At z = 1: ez^ is analytic and e^1 ̸= 0. Thus g(z) = e
(^1) + O(z − 1) (z − 1)^4 ·^
1 (1−(z−1)) — the factor (z − 2) is nonzero at 1. So z = 1 is a pole of order 4. Principal part coefficient of (z − 1)−^4 equals e · (^) −^11 times (careful expansion for exact principal part), but main classification is order 4.
At z = 2: denominator has simple zero; numerator e^2 ̸= 0 so z = 2 is a simple pole (order 1).
Example F.2.3. (Pole via reciprocal-zero test) Consider
h(z) = 1 sin^3 z
at z = 0.
We know sin z = z − z 63 + O(z^5 ). Thus
sin^3 z = z^3
1 − z 62 + O(z^4 )
= z^3
1 + O(z^2 )
So h(z) = z−^3 (1 + O(z^2 ))−^1 = z−^3 + O(z−^1 ). Therefore z = 0 is a pole of order 3. The residue is the coefficient of z−^1 in the full expansion (requires more terms if needed).
contains other singularities. Trap illustrated: Attempting to expand about 0 fails; one must treat 0 as accumulation point.
Example F.4.2. (Essential vs accumulation subtlety)
g(z) =
n=
z − 1 /n.
Analysis. The function is meromorphic on C \ { 1 /n : n ∈ N} with simple poles at 1 /n, and these poles accumulate at 0. Therefore 0 is not an isolated singularity. The series defining g may diverge near large sets — the point 0 is a natural boundary for meromorphic continuation.
Example F.4.3. (Singularity at infinity) Consider the entire function f (z) = ez^. Classify the singularity at z = ∞.
Solution. Let w = 1/z. Consider F (w) = f (1/w) = exp(1/w). At w = 0, this has
Laurent expansion
n=
n!
w−n^ (infinitely many negative powers) so w = 0 is essential;
hence z = ∞ is an essential singularity of ez^. (Contrast: a polynomial has a pole at infinity of order equal to its degree.)
When asked to classify a singularity at z 0 :
n=−∞ anzn^ is its Laurent expansion around z = 0. Which of the following statements are true? (a) If a− 1 = 0 and a−n = 0 for all n > 1, then z = 0 is removable. (b) If a− 2 ̸= 0 and a−n = 0 for n > 2, then z = 0 is a pole of order 2. (c) If infinitely many a−n ̸= 0, then z = 0 is an essential singularity. (d) If f (z) has zeros accumulating at z = 0, then z = 0 is non-isolated.
(a) z = 1 is a pole. (b) The order of the pole is 3. (c) Laurent expansion around z = 1 has negative powers up to 1/(z − 1)^3. (d) z = −1 is a removable singularity.
(a) Poles occur at z = n, n ∈ Z. (b) All poles are simple. (c) Accumulation of poles at infinity implies ∞ is non-isolated.
(c) Let f (z) = log(sin z). Identify true statements: i. z = nπ, n ∈ Z are branch points. ii. z = nπ are isolated singularities. iii. f (z) requires branch cuts along real axis to define a single-valued branch. iv. z = π/2 is a singularity. (d) Consider f (z) = (^) z log^1 z. Determine correct statements:
i. z = 0 is a non-isolated singularity. ii. z = 0 is an essential singularity. iii. z = 1 is a regular point. iv. Laurent expansion exists in 0 < |z| < 1. (e) Let f (z) = e^1 /(z^2 −1). Which statements are correct? i. z = 1 and z = −1 are essential singularities. ii. Laurent series around z = 0 has only finitely many negative powers. iii. z = 0 is an ordinary point. iv. The function is unbounded near z = 1. (f) Consider f (z) = log((z−z2)−1) 2. Identify correct statements: i. z = 1 is a branch point. ii. z = 2 is a pole of order 2. iii. z = 0 is a removable singularity. iv. Branch cut needed from z = 1 to infinity.
(g) Let f (z) = (^1) z + (^) log^1 z. Determine correct statements: i. z = 0 is a non-isolated singularity. ii. z = 0 is an essential singularity. iii. Laurent expansion exists in punctured neighborhood of z = 0. iv. f (z) is bounded near z = 0.
(h) Consider f (z) = sin
log^1 z
. Which statements are correct? i. z = 1 is an essential singularity. ii. z = 0 is a branch point. iii. Laurent series around z = 1 has infinitely many negative powers. iv. f (z) is analytic in a neighborhood of z = 1. (i) Let f (z) = log