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Cauchy’s Integral Formula
Statement
Let f (z) be analytic in a simply connected domain D containing the simple closed positively oriented contour C. If z 0 is a point inside C, then
f (z 0 ) =
2 πi
Z
C
f (z) z − z 0
dz.
—
Proof (Step by Step)
- Let ϵ > 0. Construct a small circle Cϵ : |z − z 0 | = ϵ lying inside C.
- For z on Cϵ, write:
f (z) z − z 0
f (z 0 ) z − z 0
f (z) − f (z 0 ) z − z 0
- Therefore, Z
Cϵ
f (z) z − z 0
dz = f (z 0 )
Z
Cϵ
z − z 0
dz +
Z
Cϵ
f (z) − f (z 0 ) z − z 0
dz.
- The first integral is f (z 0 ) · (2πi) because
R
Cϵ
dz z−z 0 = 2πi.
- The second integral tends to 0 as ϵ → 0, since |f (z)−f (z 0 )| < δ (continuity of f ).
- Thus, (^) Z
C
f (z) z − z 0
dz = 2πi f (z 0 ).
f (z 0 ) =
2 πi
Z
C
f (z) z − z 0
dz
—
Hidden / Key Insights
- The domain must be simply connected; otherwise, path dependence may occur (e.g., f (z) = 1/z in C \ { 0 }).
- This theorem shows that values of an analytic function inside a do- main are completely determined by its values on the boundary.
- It also gives direct formulas for derivatives:
f (n)(z 0 ) =
n! 2 πi
Z
C
f (z) (z − z 0 )n+^
dz.
- It is a fundamental result connecting boundary integrals with function values inside.
—
Solved MSQ Examples
Q1. Evaluate
R
|z|=
z^2 z− 1 dz. Solution: Here f (z) = z^2 , z 0 = 1, inside |z| = 2. Z
|z|=
z^2 z − 1
dz = 2πi · f (1) = 2πi · 12 = 2πi.
Q2. Evaluate
R
|z|=
ez z− 2 dz. Solution: Here f (z) = ez^ , z 0 = 2, inside |z| = 3. Z
|z|=
ez z − 2
dz = 2πi · f (2) = 2πi · e^2.
Q3. Evaluate
R
|z|=
sin z z dz. Solution: Here f (z) = sin z, z 0 = 0, inside |z| = 1. Z
|z|=
sin z z
dz = 2πi · f (0) = 2πi · 0 = 0.
Q4. Evaluate
R
|z|=
cos z z− 3 dz. Solution: Here z 0 = 3, but it is outside |z| = 2. So integral = 0 (since f is analytic inside and on contour). — Q5. Evaluate
R
|z− 2 |=
1 z^2 +1 dz. Solution: Poles are at z = i, −i. Neither lies inside |z − 2 | = 1. So integral = 0. —
Uniform Limits and Contour Integration
Solved Examples
Example 1
Given: fn(z) =
zn n
, C : |z| = 1.
Analysis: For |z| = 1, |fn(z)| ≤ 1 /n, so fn → 0 uniformly on C. Conclusion:
n^ lim→∞
Z
C
fn(z) dz =
Z
C
0 dz = 0.
Example 2
Given: fn(z) =
z n , C : |z| = 2.
Analysis: On C, |fn(z)| ≤ 2 /n → 0 uniformly. Also
R
C z dz^ = 0, so each integral is 0. Conclusion:
lim n→∞
Z
C
fn(z) dz = 0 =
Z
C
0 dz.
Example 3
Given: fn(z) = zn, C : |z| = 1. Analysis: On |z| = 1, fn(eiθ^ ) = einθ^ does not converge uniformly (it does not converge at all pointwise for general θ). The theorem does not apply. Note nevertheless that
R
|z|=1 z
n (^) dz = 0 for each n ≥ 1, but interchange of limit is not
justified by uniform convergence. Conclusion: Uniform convergence fails; cannot interchange limit in general.
Example 4
Given: fn(z) =
n + z
, C : |z| = 1.
Analysis: For |z| = 1 and n ≥ 1, |n + z| ≥ n − 1, so |fn(z)| ≤ 1 /(n − 1). Hence fn → 0 uniformly on C. Conclusion:
lim n→∞
Z
C
fn(z) dz =
Z
C
0 dz = 0.
Example 5
Given: fn(z) =
z z + n , C : |z| = 2.
Analysis: For z on C we have |z + n| ≥ n − 2, so
z z + n
n − 2
uniformly for n > 2. Thus fn → 0 uniformly. Conclusion:
lim n→∞
Z
C
fn(z) dz =
Z
C
0 dz = 0.
A Cautionary Counterexample (why uniform is
needed)
Let fn(z) = zn^ on C : |z| = 1. Pointwise fn(eiθ^ ) does not converge for generic θ, and uniform convergence fails. Although
R
C z
n (^) dz = 0 for each n ≥ 1, you
cannot justify
lim n→∞
Z
C
zn^ dz =
Z
C
lim n→∞ zn^ dz
by the theorem because the limit function does not exist (or is not continuous) on C.
Morera’s Theorem — Statement, Proof, Deep
Meaning and Applications
Statement
Let f (z) be continuous on a domain D ⊆ C. If I
C
f (z) dz = 0 for every simple closed contour C ⊂ D,
then f (z) is analytic in D.
Proof (Step by Step)
- Fix a point z 0 ∈ D. For any z ∈ D, define
F (z) =
Z
γ
f (ζ) dζ,
where γ is any path in D from z 0 to z.
Q
f (z) = z
Which are true? (A) f (z) is continuous everywhere, (B)
H
C f^ (z)^ dz^ = 0 for all closed contours, (C) f (z) is analytic, (D) f (z) violates Cauchy–Riemann equations. Solution: (A), (D) true; (B), (C) false. —
Q
f (z) =
z^2 , z ̸= 0, 0 , z = 0
Which are true? (A) f (z) is continuous everywhere, (B)
H
C f^ (z)^ dz^ = 0, (C) f (z) is analytic everywhere, (D) Morera’s theorem applies. Solution: All (A), (B), (C), (D) are true. —
Q
f (z) =
z
, z ∈ C \ { 0 }
Which are true? (A) f (z) is continuous on its domain, (B)
H
|z|=1 f^ (z)^ dz^ = 0, (C) f (z) is analytic in C \ { 0 }, (D) Morera’s theorem cannot apply here. Solution: (A), (C), (D) true; (B) false. —
Q
f (z) =
ez^ , z ̸= 0, 0 , z = 0
Which are true? (A) f (z) is discontinuous at 0, (B)
H
C f^ (z)^ dz^ = 0 for all contours, (C) f (z) is analytic on C \ { 0 }, (D) Morera’s theorem fails due to discontinuity. Solution: (A), (C), (D) true; (B) false.
Antiderivative Theorem / Integral of Analytic
Functions
Statement
Let f (z) be analytic in a simply connected domain D. Fix z 0 ∈ D. Define
F (z) =
Z (^) z
z 0
f (ζ) dζ,
where the integral is taken along any path in D from z 0 to z. Then F (z) is analytic in D and F ′(z) = f (z).
Proof (Step by Step)
- Since D is simply connected and f is analytic, the integral is path- independent by Cauchy–Goursat theorem.
- For h → 0, consider the difference quotient:
F (z + h) − F (z) h
h
Z (^) z+h
z
f (ζ) dζ.
- By continuity of f (analytic =⇒ continuous), as h → 0, the integral behaves like f (z) · h.
- Hence, lim h→ 0
F (z + h) − F (z) h
= f (z). Therefore, F ′(z) = f (z) and F (z) is analytic. □
Deep Meaning and Insights
- Every analytic function in a simply connected domain has an antideriva- tive.
- Path independence is guaranteed by simple connectivity.
- Analyticity ensures smoothness, so F (z) is differentiable.
- Applications: Compute integrals via antiderivatives, prove Cauchy inte- gral formula, solve potential theory problems.
Q
f (z) =
1 + z^2
, D = C \ {i, −i}, z 0 = 0
F(z) =
R (^) z 0 1 /(1 +^ ζ
(^2) )dζ. Which are correct? (A) F (z) = arctan(z) (B) F ′(z) = 1/(1 + z^2 ) (C) F (z) is analytic in D (D) F (z) = ln(1 + z^2 ) Solution: (A), (B), (C) correct; (D) false.
Uniform Limit of Analytic Functions is Analytic
Statement
Let {fn(z)} be a sequence of analytic functions on a domain D. If fn(z) con- verges uniformly on every compact subset of D to a function f (z), then f (z) is analytic in D. Moreover, the derivatives converge uniformly on compact subsets: f (^) n′(z) → f ′(z).
Proof (Step by Step)
- Let K ⊂ D be any compact subset. Since {fn} converges uniformly on K, the limit function f (z) is continuous on K.
- For each fn, by Cauchy’s integral formula:
fn(z) =
2 πi
I
C
fn(ζ) ζ − z
dζ,
where C is any simple closed contour inside K.
- Uniform convergence allows passing the limit under the integral sign:
f (z) = lim n→∞ fn(z) =
2 πi
I
C
f (ζ) ζ − z
dζ.
- By Cauchy’s formula, f (z) is analytic in D, and derivatives converge sim- ilarly. □
Deep Meaning and Insights
- Uniform convergence preserves analyticity.
- Pointwise convergence is not enough; analytic property may fail.
- Applications: Construct new analytic functions as limits (series expan- sions, sequences of polynomials).
- Mechanism: Use Cauchy integral formula + uniform convergence to guarantee limit function is analytic.
5 Solved MSQ Examples
Q
fn(z) =
zn 1 + n^2
, D = C
Which are correct? (A) fn → 0 uniformly on compact sets (B) Limit is analytic (C) Converges pointwise only (D) Limit is discontinuous Solution: - On any compact disk |z| ≤ R, |zn/(1 + n^2 )| ≤ Rn/(1 + n^2 ) → 0 uniformly. - Limit is 0 (analytic). - Correct: (A), (B); (C), (D) false. —
Q
fn(z) =
z n
n , D = C
(A) fn → ez^ uniformly on compact sets (B) Limit is analytic (C) Converges pointwise only (D) Limit not analytic Solution: - Standard limit: (1 + z/n)n^ → ez^ uniformly on compact sets. - Limit ez^ is entire (analytic). - Correct: (A), (B); (C), (D) false. —
Q
fn(z) =
z 1 + nz^2
, D = C
(A) fn → 0 uniformly on compact sets (B) Limit analytic
Proof
- Start with Cauchy’s integral formula for the n-th derivative:
f (n)(z 0 ) =
n! 2 πi
I
C
f (z) (z − z 0 )n+^
dz
- Take modulus on both sides:
|f (n)(z 0 )| ≤
n! 2 π
I
C
|f (z)| |z − z 0 |n+^ |dz|
- On the circle C, |z − z 0 | = r and |f (z)| ≤ M , and the length of C is 2πr:
|f (n)(z 0 )| ≤ n! 2 π
M
rn+^
· 2 πr = n!M rn
- Hence proved. □
Deep Meaning and Insights
- Provides an upper bound for the derivatives of an analytic function.
- Shows how the size of a function on a circle controls its derivatives at the center.
- Useful for: Liouville’s theorem, uniqueness of analytic continuation, and estimating derivatives in series expansions.
- Mechanism: Derived directly from Cauchy’s integral formula and the mod- ulus inequality.
5 Solved MSQ Examples
Q
f (z) = ez^ , |z| = 1 =⇒ M = e
Which of the following are correct? (A) |f ′(0)| ≤ 1 (B) |f ′(0)| ≤ e (C) |f ′(0)| = e (D) |f ′(0)| ≤ 2 e Solution:
|f ′(0)| ≤
1! · e 11
= e
- Correct: (B), (C) (both indicate same bound). - Incorrect: (A), (D)
Q
f (z) = sin(z), |z| = 2 =⇒ M = sinh(2) ≈ 3. 626
Find upper bound for |f ′′(0)|. (A) 3.626 (B) 1.813 (C) 2*3.626/4 (D) 7. Solution:
- Here n = 2, r = 2, M = 3. 626
|f ′′(0)| ≤
- Correct: (B), (C) - Incorrect: (A), (D)
Q
f (z) = z^3 + 2z, |z| = 1 =⇒ M = |f (z)|max = 3
Find upper bound for |f ′′′(0)|. (A) 18 (B) 6*3 (C) 3 (D) 1 Solution:
|f ′′′(0)| ≤
- Correct: (A), (B) - Incorrect: (C), (D)
Q
f (z) =
1 + z , |z| = 1 =⇒ M = max |z|= |f (z)| =
Find upper bound for |f ′(0)|. (A) 1 (B) 0.5 (C) 2 (D) 0. Solution:
|f ′(0)| ≤
- Correct: (B) - Incorrect: (A), (C), (D)
Statement
If f (z) is entire (analytic on C) and bounded, then f (z) is constant.
Formally:
- Entire function: f (z) analytic everywhere in C
- Bounded: ∃M > 0 such that |f (z)| ≤ M, ∀z ∈ C
- Conclusion: f ′(z) = 0 =⇒ f (z) = constant
Hidden Concepts / Deep Meaning
- Boundedness + Entirety ⇒ Constancy: Analytic everywhere is not enough; boundedness restricts growth. Non-constant entire functions grow without bound.
- Derived from Cauchy’s inequality:
|f ′(0)| ≤
M
R
=⇒ |f ′(0)| = 0 as R → ∞
- Implication: No bounded non-constant entire function exists. Used to prove the Fundamental Theorem of Algebra:
P (z) polynomial =⇒ 1 /P (z) cannot be bounded if no roots exist.
- Connection with growth: Liouville’s theorem is the simplest case of growth restrictions of entire functions: entire + bounded ⇒ growth = 0.
Q
Let f (z) be entire and |f (z)| ≤ 5 for all z ∈ C. Then: (A) f (z) = 0 (B) f (z) is constant (C) f (z) may be non-constant (D) f (z) is linear Solution: - By Liouville’s theorem, any bounded entire function is constant.
Q
Which of the following functions are constant by Liouville’s theorem? (A) f (z) = ez (B) f (z) = sin(z) (C) f (z) = 2 (D) f (z) = z^2 Solution: - Only f (z) = 2 is bounded and entire. - Others are unbounded on C. - Correct: (C)
Q
Let f (z) be entire and |f (z)| ≤ |z|^2 + 1. What can be said about f (z)? (A) Constant (B) Linear (C) Quadratic (D) Cannot be determined Solution: - Here f (z) is not bounded (grows as |z|^2 ), so Liouville’s theorem does not apply. - Cannot conclude f is constant. - Correct: (D)
Q
Which of the following is a consequence of Liouville’s theorem? (A) Fundamental Theorem of Algebra (B) Cauchy’s inequality (C) Morera’s theorem (D) Schwarz lemma Solution: - Liouville’s theorem implies any non-constant polynomial must have a root in C, which is the Fundamental Theorem of Algebra. - Correct: (A)
Q
If f (z) is entire and |f (z)| ≤ 100 for all z ∈ C, then: (A) f (z) = 0 (B) f (z) is constant (C) f (z) is linear (D) f (z) can be non-constant Solution: - By Liouville’s theorem, bounded entire functions are constant.
Analytic Maps from C to Unit Disk
5 Solved MCQs
Q
Let f : C → ∆ be analytic. Then: (A) f (z) = 0 (B) f (z) is constant (C) f (z) may be non-constant (D) f (z) is linear Solution: By Liouville’s theorem, bounded entire function must be constant. Correct: (B)
Q
Which of the following functions is analytic from C to ∆? (A) f (z) = 1/ 2 (B) f (z) = ez^ / 2 (C) f (z) = tanh(z) (D) f (z) = z Solution: Only f (z) = 1/2 is bounded by 1 and entire. Correct: (A)
Q
Is there a bijective analytic map from ∆ onto C? (A) Yes (B) No (C) Only if linear (D) Only if exponential Solution: Analytic map from C to ∆ is constant → cannot be bijective. Correct: (B)
Q
Let f : C → C be analytic and bounded. Then: (A) f (z) is polynomial of degree 1 (B) f (z) is entire and bounded → constant (C) f (z) can grow at infinity (D) f (z) is transcendental Solution: Liouville’s theorem: bounded entire function → constant. Correct: (B)
Q
Let f : C → ∆ be entire and non-constant. Which is true? (A) f (z) exists
(B) Impossible (C) f (z) is linear (D) f (z) is polynomial Solution: By Liouville, no non-constant bounded entire function exists. Cor- rect: (B)
Generalized Liouville’s Theorem /
Picard’s Little Theorem
Statement
Every non-constant entire function f : C → C takes values arbitrarily close to every complex number, possibly missing at most one exceptional value. Example: ez^ misses 0 but comes arbitrarily close to all other complex num- bers. Polynomials attain all complex numbers.
Proof Idea
- Let f be non-constant entire.
- Assume f avoids some value w 0 ∈ C. Then g(z) = (^) f (z)^1 −w 0 is entire and bounded.
- By Liouville’s theorem, g(z) is constant, leading to a contradiction.
- Hence f (z) must get arbitrarily close to every w ∈ C except at most one value.
Conclusion: Non-constant entire functions are dense in C.
Hidden Concepts / Deep Meaning
- Bounded Entire Function ⇒ Constancy: Any bounded entire function must be constant (Liouville theorem).
- Exceptional Value: Some entire functions may never attain one value (like ez^ misses 0).
- Density in C: Non-constant entire functions get arbitrarily close to almost all complex numbers.
- Function-theoretic Implication: Entire functions cannot avoid two or more values without being constant.
