Analytic Functions

1. Cauchy–Riemann equations, harmonic functions

Exam focus: CR equations alone do not give holomorphy — you need them plus continuity of the partials (or real-differentiability). The standard trap is a function satisfying CR only at the origin.

Lecture 3.4 - Cauchy Riemann equations

NPTEL · Complex Analysis (Pranav Haridas, KSoM)

Note carefully what CR alone does and does not give you.

Lecture - 4.1 - Harmonic functions

NPTEL · Complex Analysis (Pranav Haridas, KSoM)

Harmonic conjugates and where simple connectedness is needed.

Cauchy–Riemann — what they do and do not give you free

The equations

f = u + iv is complex differentiable at $z_{0}$ iff f is real-differentiable there and $u_{x} = v_{y}, u_{y} = - v_{x}$ . Then $f^{'} = u_{x} +$ iv $_{x}$ .

Hypothesis	Conclusion
CR hold at a single point + real-differentiable there	f′ exists at that point
CR hold on an open set and partials are continuous	f is holomorphic there (Goursat: continuity is automatic)
CR hold on an open set, partials merely exist	not enough

**The classic spoiler: $* * f (z) = (\overset{z}{ˉ})^{2} / z$ for z ≠ 0, f(0) = 0 satisfies the CR equations at 0 but is not differentiable at 0 (the limit depends on the direction of approach).

Consequences worth quoting

f holomorphic with f′ ≡ 0 on a domain (connected open set) ⇒ f constant. Connectedness is essential.
Any of these forces a holomorphic f on a domain to be constant: |f| constant, Re f constant, Im f constant, arg f constant, f̄ holomorphic, f real-valued.
u and v are **harmonic $* * (Δ u = 0)$ and are harmonic conjugates; the level curves $u = c_{1}$ and $v = c_{2}$ meet orthogonally where $\nabla u \neq = 0$ . $- \nabla u \cdot \nabla v = 0$ for holomorphic $f : u_{x} v_{x} + u_{y} v_{y} = u_{x} (- u_{y}) + u_{y} u_{x} = 0$ .
A harmonic conjugate exists on any simply connected domain. On an annulus it can fail: u = log|z| has no single-valued conjugate on $C$ ∖{0}.

Anti-holomorphic traps

If f is entire then:

z ↦ conj(f(z̄)) is entire (conjugate the coefficients);
z ↦ f(z̄) and z ↦ conj(f(z)) are anti-holomorphic — entire only when f is constant.

Key takeaways

CR at a point ≠ holomorphic at a point; you need real-differentiability too.
Any "constant modulus / constant real part / conjugate also holomorphic" hypothesis collapses f to a constant.
conj(f(z̄)) is the one conjugation that preserves holomorphy.

Part B2023 Decembershow ▾

Let

f : C \to C

be a real-differentiable function and define u(x, y) = Re f(x + iy), v(x, y) = Im f(x + iy). Let

\nabla u = (u_{x}, u_{y})

denote the gradient. Which one of the following is necessarily true?

A.For $c_{1}, c_{2} \in C$ , the level curves $u = c_{1}$ and $v = c_{2}$ are orthogonal wherever they intersect.
B. $\nabla u \cdot \nabla v = 0$ at every point.
C.If f is an entire function, then $\nabla u \cdot \nabla v = 0$ at every point.✓
D.If $\nabla u \cdot \nabla v = 0$ at every point, then f is an entire function.

Solution

For entire f the Cauchy–Riemann equations give $\nabla u \cdot \nabla v = u_{x} v_{x} + u_{y} v_{y} = u_{x} (- u_{y}) + u_{y} u_{x} = 0$ . Without holomorphy there is no such relation, and orthogonal gradients alone do not force CR (e.g. f = z̄ has $\nabla u \cdot \nabla v = 0)$ .

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Part B2023 Decembershow ▾

Previous-year question — unlock to view

Part C2023 Juneshow ▾

Let f(z) be an entire function on

C

. Which of the following statements are true?

A.f(z̄) is an entire function.
B.conj(f(z)) is an entire function.
C.conj(f(z̄)) is an entire function.✓
D.conj(f(z̄)) + f(z̄) is an entire function.

Solution

If $f = \sum a_{n} z^{n}$ then conj $(f (\overset{z}{ˉ})) = \sum$ ā $_{n} z^{n}$ is entire. z ↦ f(z̄) and z ↦ conj(f(z)) are anti-holomorphic (entire only if f is constant). (4) is entire plus non-entire ⇒ not entire in general.

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Part Bshow ▾

The function f(z) = |z|

^{2}

is complex differentiable

A.nowhere
B.only at z = 0✓
C.everywhere
D.exactly on the real axis

Solution

$f = x^{2} + y^{2}$ has $u = x^{2} + y^{2}, v = 0$ . Cauchy–Riemann requires 2x = 0 and 2y = 0, holding only at the origin — and there the difference quotient (z̄) does tend to 0, so f is differentiable at 0 alone (and holomorphic nowhere).

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Part Cshow ▾

Let f(z) = u(x, y) + iv(x, y) be defined on

C

. Which of the following are true?

A.If f is holomorphic on $C$ , then u and v are harmonic.✓
B.If u and v satisfy the Cauchy–Riemann equations at a single point, then f is differentiable at that point.
C.f(z) = z̄ satisfies the Cauchy–Riemann equations nowhere.✓
D.If f is holomorphic and real-valued on all of $C$ , then f is constant.✓

Solution

(1) Differentiating the CR equations once more gives u_xx + u_yy = v_yx − v_xy = 0, using the smoothness of holomorphic functions. (2) fails without continuity of the partials — $f (z) = exp (- 1/ z^{4}), f (0) = 0$ satisfies CR at the origin without being differentiable there. (3) For z̄: u = x, v = −y gives u_x = 1 ≠ −1 = v_y at every point. (4) Real-valued forces v ≡ 0, then CR forces u_x = u_y = 0.

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2. Power series and analyticity

Exam focus: Holomorphic ⇔ analytic ⇔ locally a convergent power series — an equivalence with no real-analysis analogue. Use the identity theorem to force f ≡ g from agreement on a set with a limit point *inside* the domain.

Lecture 3.1- Power series

NPTEL · Complex Analysis (Pranav Haridas, KSoM)

Lecture 3.2 - Differentiation of Power series

NPTEL · Complex Analysis (Pranav Haridas, KSoM)

Holomorphic ⇔ analytic — the equivalence with no real analogue.

Analytic Functions

1. Cauchy–Riemann equations, harmonic functions

Cauchy–Riemann: satisfied at a point, holomorphic nowhereinteractivefree

Cauchy–Riemann — what they do and do not give you free

The equations

Consequences worth quoting

Anti-holomorphic traps

Key takeaways

2. Power series and analyticity

Why the radius of convergence is set by the nearest singularityinteractive

Analyticity, the identity theorem and zeros

The great equivalence

Cauchy estimates and their children