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If $f=u+iv$ is a holomorphic function on a connected domain andthen the Cauchy-Riemann equations force both components of $f^{\prime }$ to vanish. Hence $f$ is constant.

If a nonconstant complex polynomial were zero-free, its reciprocal would be an entire bounded function. The real part of its value along planar Brownian motion is a bounded martingale, hence converges almost surely. If that real part were nonconstant, continuity would give two disks with separated value ranges, and recurrence of planar Brownian motion would force alternate visits to both at arbitrarily large times, contradicting convergence. It is therefore constant; the Cauchy-Riemann equations then make the reciprocal constant, a contradiction.

The complex derivative exists when this limit has the same finite value for increments $h$ approaching zero from every complex direction. For $f=u+iv$, real and imaginary increments give the necessary Cauchy-Riemann equations. Existence of those partial derivatives at one point alone is not sufficient for complex differentiability at a point; continuous partial derivatives in a neighbourhood, together with the Cauchy-Riemann equations there, give a standard sufficient condition for $f$ to be a holomorphic function.

A complex-valued function is complex differentiable at $z_0$ ifexists. The same limit must hold for every approach, including curved paths. Equivalently,The error divided by $|h|$ tends to zero. Thus the derivative is a single complex multiplier, not an arbitrary real linear map of the plane. If the real coordinate map is differentiable, complex differentiability is equivalent to its satisfying the Cauchy-Riemann equations at that point.

The simplest failure is complex conjugation, $f(z)=\overline{z}$. At every $z_0$ its increment quotient is $\overline{h}/h$. For real $h=t\ne 0$ it is $1$, while for $h=it$ it is $-1$. More generally, $h=t{e}^{i\theta }$ gives $e^{-2i\theta }$. Although complex conjugation is a smooth real reflection and has every real directional derivative, these incompatible quotients rule out a complex derivative everywhere.

Agreement on two axes is necessary, but does not by itself establish a limit over the whole complex plane. The Cauchy-Riemann equations need the real differentiability condition above, or a sufficient regularity condition such as continuous partial derivatives near the point. There is a second distinction: $f(z)=|z{|}^2$ has complex derivative zero at zero because its quotient there is $\overline{h}\to 0$, but its Cauchy-Riemann equations fail at every nonzero point. It is complex differentiable at zero without being a holomorphic function on any neighbourhood of zero.

At zero, $[f(h)-f(0)]/h\to 1$, so the complex derivative exists. At a nonzero point, write $r=|z|$. The Cauchy-Riemann equations for $x/(1+r)$ and $y/(1+r)$ force $x^2=y^2$ and $xy=0$, impossible when $r>0$. Hence the function is complex differentiable only at zero and is holomorphic on no nonempty open subset. A complex derivative at one point gives no neighborhood of complex differentiability.

For an incompressible irrotational two-dimensional potential flow, an analytic complex potential of a two-dimensional potential flow combines the velocity potential $\Phi$ and streamfunction $\Psi$. The Cauchy-Riemann equations give $u={\Phi }_x={\Psi }_y$ and $v={\Phi }_y=-{\Psi }_x$, hence $W^{\prime }(z)=u-iv$. Curves of constant $\Psi$ are streamlines. An analytic coordinate change $z=f(\zeta )$ transforms the complex velocity by $dW/dz=(dW/d\zeta )/f^{\prime }(\zeta )$ wherever $f^{\prime }\ne 0$.

For a complex potential of a two-dimensional potential flow $W=\varphi +i\psi$ in the coordinate $z=x+iy$, the Cauchy-Riemann equations give $W^{\prime }={\varphi }_x-i{\varphi }_y=u-iv$. A material point with physical velocity $(u,v)$ therefore satisfies $\dot{z}=\overline{W^{\prime }(z)}$. Complex conjugation matters when deriving a pathline from a complex potential of a two-dimensional potential flow.

If $u+iv$ and $p+iq$ are holomorphic functions, with the image of the first inside the domain of the second, then $p(u,v)+iq(u,v)$ is holomorphic by the chain rule. Directly, their Cauchy-Riemann equations imply $[p(u,v){]}_x=[q(u,v){]}_y$ and $[p(u,v){]}_y=-[q(u,v){]}_x$. Differentiating these relations proves harmonicity of both parts. No nonvanishing derivative hypothesis is required.

A holomorphic function with constant modulus on a connected domain is constant. If $h=u+iv$ and $u^2+v^2=c^2>0$, differentiation and the Cauchy-Riemann equations give $u{u}_x+v{v}_x=0$ and $-u{v}_x+v{u}_x=0$. Their coefficient determinant is $-c^2$, so $u_x=v_x=0$, and all first derivatives vanish. When $c=0$, the conclusion is immediate. Consequently two nonvanishing holomorphic functions with equal moduli differ by a single constant phase on a connected domain.

A distribution $u$ on a plane open set with ${\partial }_{\bar{z}}u=0$ is represented by a holomorphic function. Indeed $\Delta u=4{\partial }_z{\partial }_{\bar{z}}u=0$, and elliptic regularity makes $u$ smooth. For a direct proof, mollify locally with a radial mollifier ${\eta }_{\epsilon }$. The resulting smooth functions are harmonic. Their mean value property gives $(u\ast {\eta }_{\epsilon })\ast {\eta }_r=u\ast {\eta }_{\epsilon }$ on a smaller ball for fixed small $r$. Passing to the weak convergence of distributions limit gives $u\ast {\eta }_r=u$ there. This proves smoothness without assuming local integrability beforehand. The ordinary Cauchy-Riemann equations now prove holomorphy.

The equation ${\partial }_{z}u=0$ similarly characterizes antiholomorphic functions. On the whole plane, the homogeneous solutions are respectively arbitrary entire functions $h(z)$ and the functions $h(\bar{z})$ with $h$ entire. Consequently any two fundamental solutions for one of these Wirtinger derivatives differ by the corresponding homogeneous solution.

If $f:U\to V$ is a holomorphic function between planar open sets and $\psi :V\to \mathbb{R}$ is a harmonic function, then $\psi \circ f$ is harmonic. In fact the chain rule and the Cauchy-Riemann equations, with $f=u+iv$, giveHere the two gradients are orthogonal and have equal length, while $u,v$ are harmonic. The formula holds even where $f^{\prime }=0$. A biholomorphism therefore induces inverse correspondences between harmonic functions by function composition with it and its inverse.

For a smooth nonconstant holomorphic disk $h$ in ${\mathbb{C}}^n$ with boundary on a Lagrangian submanifold $L$, its symplectic area is ${\int }_{D^2}|{\partial }_{s}h{|}^{2}dsdt>0$. The Cauchy-Riemann equations give ${\partial }_{t}h=J{\partial }_{s}h$, so this is the pullback of the standard symplectic form. The Stokes theorem identifies the area with the period of a primitive along the boundary loop. Thus the Liouville class of a Lagrangian submanifold evaluates nontrivially on this loop and cannot vanish.

A complex-valued function $f:U\to \mathbb{C}$ is holomorphic on an open set $U$ when it has a complex derivative at every point of $U$. Requiring a derivative throughout a neighbourhood is much stronger than complex differentiability at a point. It forces locally convergent Taylor series and derivatives of every order, as follows from the Cauchy integral formula and Cauchy derivative formula.

To see the restriction in real coordinates, write $z=x+iy$ and $f(z)=u(x,y)+iv(x,y)$. The real and imaginary increment quotients must agree, giving the Cauchy-Riemann equationsConversely, if $u,v$ have continuous partial derivatives on $U$ and satisfy the Cauchy-Riemann equations there, then $f$ is a holomorphic function. The equations at a lone point, without a differentiability hypothesis for the real coordinate map, are insufficient. Their role is to force the Jacobian matrix to have the special formSuch a matrix represents multiplication by a complex number; it has only two free real entries. ^[ref]

For $f(z)=z^2$, $u=x^2-y^2$ and $v=2xy$, so $u_x=v_y=2x$ and $u_y=-v_x=-2y$. Directly,At $z_0=1+i$, multiplication by $2+2i$ scales lengths by its modulus $2\sqrt{2}$ and rotates by its argument $\pi /4$. A horizontal increment $h=0.01$ gives the exact displacement $0.0201+0.02i$; a vertical increment $h=0.01i$ gives $-0.0201+0.02i$. The first-order predictions $0.02+0.02i$ and $-0.02+0.02i$ are perpendicular, and the errors have modulus $10^{-4}$.

At every nonzero $z_0$, this local rotation and scaling makes $z^2$ a local conformal map. At zero, the first-order map vanishes: $r{e}^{i\theta }$ becomes $r^2{e}^{2i\theta }$, so the argument doubles and the origin is a critical point. Globally $z$ and $-z$ have the same square. Local preservation of angles therefore coexists with global folding, a useful warning when using holomorphic functions as coordinate changes.

Let $g$ be a holomorphic function on an open set $\Omega \subseteq \mathbb{C}$. Among continuously differentiable functions, all solutions of ${\partial }_{\bar{z}}f=g$ arewhere $h$ is an arbitrary holomorphic function on $\Omega$. The Wirtinger derivatives satisfy ${\partial }_{\bar{z}}\bar{z}=1$ and ${\partial }_{\bar{z}}g=0$, so the product rule verifies the particular solution $\bar{z}g(z)$. Subtracting it from any other solution leaves a continuously differentiable function with vanishing Wirtinger derivative, hence a holomorphic function by the Cauchy-Riemann equations. Conversely adding such a holomorphic function changes no prescribed derivative. For $g(z)=z$ on the complex plane, the solutions are $|z{|}^2+h(z)$ with $h$ any entire function.

Write a holomorphic function as $f=u+iv$. The Cauchy-Riemann equations make $u,v$ harmonic functions and give $u_x=v_y$, $u_y=-v_x$. The product rule therefore gives $\Delta (u^2+v^2)=2(|\nabla u{|}^2+|\nabla v{|}^2)=4|f^{\prime }{|}^2$. Thus its squared modulus is subharmonic and is harmonic only where $f^{\prime }$ vanishes.

In a supersymmetry-preserving local Wilsonian effective action, the gauge kinetic function receives only a one-loop perturbative correction. For one simple gauge factor, $\mu df/d\mu =b_0/(8{\pi }^2)$ with $b_0=3C_2(G)-{\sum }_{i}T(R_i)$. A perturbative shift of the topological angle leaves the running coefficient invariant; the Cauchy-Riemann equations then make its holomorphic beta function independent of the gauge coupling. Holomorphic regularity and spurionic R-charges exclude higher-loop Yukawa combinations, which would require antichiral couplings. The constant is fixed by one-loop matching. This does not assert one-loop exactness of a canonically normalized physical coupling, or forbid nonperturbative terms.

// Target: quantum-field-theory.bigb

For a holomorphic function $f=u+iv$, the Cauchy-Riemann equations give the displayed gradient identities. Where $f^{\prime }\ne 0$, both level sets are regular curves; their nonzero normal vectors are perpendicular, so their tangents are perpendicular as well. The qualification is necessary. At a critical point both gradients vanish and the zero level sets may have several tangent branches. For $f(z)=z^2$ at zero, $u=0$ consists of two diagonals while $v=0$ consists of the two coordinate axes; there is no general orthogonality assertion at that point.

For a holomorphic function $f=u+iv$, the Cauchy-Riemann equations imply $\nabla u\cdot \nabla v=0$ and $|\nabla u|=|\nabla v|=|f^{\prime }|$. At a point where $f^{\prime }\ne 0$, the two regular level curves therefore meet at right angles, since each tangent is perpendicular to its nonzero gradient. The noncritical condition is necessary: for $f(z)=z^2$, the zero level sets are the coordinate axes and the two diagonals, which meet at angles of $\pi /4$ at the critical point zero. Orthogonality at regular points is one geometric expression of conformality.

Writing $z=x+iy$ and $f=u+iv$, the Cauchy-Riemann equations areWith continuous second derivatives, mixed derivatives commute. ThusBoth real and imaginary parts are harmonic.

For $g(z)=\mathrm{Re}z$, the real and imaginary parts are $x$ and zero, so $u_x=1\ne 0=v_y$. For $h(z)=\bar{z}$, they are $x$ and $-y$, so $u_x=1\ne -1=v_y$. The necessary Cauchy-Riemann equations fail at every point. Hence both functions are nowhere holomorphic, indeed nowhere complex differentiable.

Write $z=x+iy$, so the real and imaginary components are $u=x^2$ and $v=y^2$. The Cauchy-Riemann equations require $u_x=v_y$, namely $2x=2y$; the other equation is $u_y=-v_x=0$. The first relation fails off the line $x=y$, so the function is not holomorphic on $\mathbb{C}$. It is complex differentiable at points of that line, but a line contains no open neighborhood on which the equations hold.