If is holomorphic on and inside a simple closed contour , then . If is inside ,
The value at an interior point is determined entirely by the boundary values.
Cauchy Theory
1. Cauchy's theorem and integral formula
Exam focus: Pick the right tool: Cauchy's theorem (no singularities inside) vs the integral formula (one simple pole) vs residues (several). Simple connectedness is what makes ∮ = 0.
Lecture - 6.2 Cauchy-Goursat theorem
NPTEL · Complex Analysis (Pranav Haridas, KSoM)
Lecture - 6.3 Cauchy's theorem
NPTEL · Complex Analysis (Pranav Haridas, KSoM)
Watch where simple connectedness enters — that is the whole point.
Lecture - 7.1 Cauchy Integral Formula
NPTEL · Complex Analysis (Pranav Haridas, KSoM)
Part Cshow ▾Let f be entire. Which of the following are true?
- A.If f is bounded then f is constant✓
- B.If Re f is bounded then f is constant✓
- C.If as |z| then f is a polynomial✓
- D.If f has no zeros then f is constant
Solution
(b) apply Liouville to e^f. (c) pole at polynomial. (d) false: eᶻ.
2. Liouville, Morera, maximum modulus principle
Exam focus: Recognise Liouville in disguise: bounded, bounded real part, bounded on a growth scale, or missing two values. Maximum modulus turns interior bounds into boundary bounds.
Lecture - 7.2 Principle of analytic continuation and Cauchy estimates
NPTEL · Complex Analysis (Pranav Haridas, KSoM)
Cauchy estimates are the engine behind Liouville.
Lecture - 7.3 Further consequences of Cauchy Integral Formula
NPTEL · Complex Analysis (Pranav Haridas, KSoM)
Liouville and the fundamental theorem of algebra.
Lecture 8.2 - Open mapping theorem
NPTEL · Complex Analysis (Pranav Haridas, KSoM)
Open mapping ⇒ maximum modulus principle.
Part C2023 Decembershow ▾Previous-year question — unlock to view
Part C2023 Decembershow ▾Previous-year question — unlock to view
Part Cshow ▾Let f be entire. Which of the following conditions, each on its own, force f to be constant?
- A.|f(z)| ≤ 5 for all .✓
- B.Re f(z) > 0 for all .✓
- C.|f(z)| ≤ for all |z| ≥ 1.
- D.f(1/n) = 0 for every positive integer n.✓
Solution
(1) Liouville directly. (2) g = 1/(f + 1) is entire with |f + 1| ≥ Re f + 1 > 1, so g is bounded, hence constant, hence f is. (4) The zeros 1/n accumulate at , so the identity theorem forces f ≡ 0 — constant. (3) does NOT force constancy: by the Cauchy estimates the growth bound makes f a polynomial of degree ≤ 1, and f(z) = z satisfies the bound while being non-constant.
Part Bshow ▾Let f be entire with |f(z)| ≥ 1 for all . Then
- A.f must be constant.✓
- B.f must be a polynomial of positive degree.
- C.f must have a zero.
- D.no such f exists.
Solution
f never vanishes, so 1/f is entire, and |1/f| ≤ 1 makes it bounded — Liouville forces 1/f, hence f, constant. Such f do exist (any constant of modulus ≥ 1), ruling out the last option.
Part Cshow ▾Let f be holomorphic on the open unit disc D and continuous on its closure, with |f(z)| = 3 for every |z| = 1. Which of the following are possible?
- A.f is constant with |f| = 3.✓
- B.f is non-constant and has no zeros in D.
- C.f is non-constant and has at least one zero in D.✓
- D.|f(0)| > 3.
Solution
Maximum modulus gives |f| ≤ 3 on D, so (4) is impossible. If f never vanishes, the minimum modulus principle (maximum modulus applied to 1/f) gives |f| ≥ 3 inside too, forcing |f| ≡ 3 and f constant — so (2) is impossible, and every non-constant example must vanish somewhere: f(z) = 3z realises (3). (1) is realised by f ≡ 3.