Cauchy–Riemann equations, harmonic functions

Why this is asked: 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.

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: $** f(z) = ({\bar{z}})^{2}/z$ 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 $** (\Delta{}u = 0)$ and are harmonic conjugates; the level curves $u = c_{1}$ and $v = c_{2}$ meet orthogonally where $\nabla{}u \ne 0$ . $- \nabla{}u \cdot \nabla{}v = 0$ for holomorphic $f: u_{x}v_{x} + u_yv_y = u_{x}(-u_y) + u_yu_{x} = 0$ .
A harmonic conjugate exists on any simply connected domain. On an annulus it can fail: u = log|z| has no single-valued conjugate on $\mathbb{C}$ ∖{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.

See it move

Cauchy–Riemann: satisfied at a point, holomorphic nowhereinteractivefree

Step 1 / 4The equations

Writing $f = u + i v$ , complex differentiability at a point requires

$u_{x} = v_{y}, u_{y} = - v_{x} .$

These are necessary. They are not, on their own, sufficient.

The trap here

“The Cauchy–Riemann equations at a point imply complex differentiability there” — false

$f(z) = ({\bar{z}})^{2}/z$ for z ≠ 0, f(0) = 0

CR hold at 0, but the difference quotient along z = t(1+i) differs from the one along the real axis.