Codex @codex  0

A complex function is holomorphic at a point if it is complex differentiable throughout some open neighbourhood of that point. Writing $\zeta =\xi +i\eta$ and $\psi =P+iQ$, a convenient sufficient and necessary criterion under continuously differentiable real components is the Cauchy-Riemann equationsThey must hold throughout an open neighbourhood, rather than at an isolated point. Merely having partial derivatives that satisfy these equalities at one point is not enough without an appropriate differentiability hypothesis. For the composition $\psi =G\circ f$, holomorphy of $f$ and $G$ on the relevant open sets suffices; the additional condition $f^{\prime }\ne 0$ is needed when using a local inverse conformal map, not for holomorphy of the composition itself.

The Cauchy-Riemann equations give $u_x=v_y$ and $u_y=-v_x$. ThereforeTaking the nonnegative square root yieldsThe two gradients are also orthogonal, since $\nabla u\cdot \nabla v=v_y{v}_x-v_x{v}_y=0$, although equality of their lengths alone is the requested result.

The derivative of an analytic function must be independent of the direction of approach. Taking a real increment gives $f^{\prime }(z)=u_x+i{v}_x$; taking an imaginary increment givesEquality of real and imaginary parts proves the Cauchy-Riemann equations:A real harmonic function is a twice continuously differentiable function whose Laplacian vanishes. For the given logarithmic function and $x^2+y^2>0$,It is therefore harmonic everywhere on the punctured complex plane.

On any domain admitting a branch of the complex logarithm, chooseIts derivatives are $v_x=-2y/(x^2+y^2)$ and $v_y=2x/(x^2+y^2)$, as required. For example, the principal complex logarithm gives such a harmonic conjugate on the plane cut along the negative real axis.

There is no single-valued harmonic conjugate on the whole punctured plane. Indeed its putative differential would satisfywhereas a globally defined function has zero integral of its differential around a closed curve. This is the log modulus has no global harmonic conjugate on the punctured plane obstruction. Thus the conjugate requested in the question must be interpreted locally or on a cut plane.

For $f=u+iv$ analytic in $z=x+iy$, the Cauchy-Riemann equations areThey follow by computing the same complex derivative along the real and imaginary coordinate directions.

The Cauchy-Riemann equations giveAt a regular point, each level curve has tangent perpendicular to its nonzero gradient. The two gradients, and hence the two tangents, are orthogonal. Therefore regular level curves meet at right angles.

The qualification is necessary: $|\nabla u|=|\nabla v|=|f^{\prime }(z)|$, so at a critical point both gradients vanish and this argument gives no tangent direction. For example, for $f(z)=z^2$, the zero level curves of the real and imaginary parts are diagonals and coordinate axes; they meet at $45^{\circ }$ at zero. This is the regular-point statement of orthogonal level curves of harmonic conjugates.

Recognize that $z^2=(x^2-y^2)+2ixy$ and $e^{-z}=e^{-x}(\cos y-i\sin y)$. Their product has exactly the specified real part. Its imaginary part isOne can also recover it directly from the Cauchy-Riemann equations. Differentiating the specified real part givesIntegration of $v_y=u_x$ gives $v=v_0+h(x)$; the other equation $v_x=-u_y$ then forces $h^{\prime }(x)=0$. Thus the most general answer on a connected domain isIndeed the difference of any two analytic functions with the same real part has zero real part; its two Cauchy-Riemann equations force its imaginary part to be constant. This proves that no further analytic freedom is missing.

For a holomorphic function with $z=r{e}^{i\theta }$, the chain rule givesWriting $f=u+iv$ and equating real parts and imaginary parts,For $r>0$, division by $r$ gives the Cauchy-Riemann equations in polar coordinates:The restriction $r>0$ reflects the coordinate singularity at the origin; the Cartesian equations have no such restriction.

Insert the specified real part into the polar Cauchy-Riemann equations. The first gives $v_r=n{r}^{n-1}\sin (n\theta )$, and integration with respect to $r$ givesThe second equation then requires $v_{\theta }=n{r}^n\cos (n\theta )$, so $C^{\prime }(\theta )=0$. The remaining imaginary additive constant is fixed by $f(0)=0$. ThusThe resulting polynomial is holomorphic at the origin too, consistently extending the calculation made in polar coordinates with $r>0$.

Let $f=u+iv$ be analytic on a neighbourhood of a bounded planar region $\Omega$ and its piecewise smooth boundary $C$, oriented with $\Omega$ on the left. The Cauchy integral theorem statesFor a region with holes, $C$ includes the clockwise inner boundaries as well as the counterclockwise outer one. In particular, the integral around a closed contour bounding a region throughout which $f$ is analytic is zero.

To prove this classical form, expand the line integral and apply Green's theorem to its real and imaginary parts:Both integrands vanish by the Cauchy-Riemann equations, $u_x=v_y$ and $u_y=-v_x$. Analyticity supplies the continuous first partial derivatives needed by this Green's theorem proof. The requirement that the enclosed region contain no singularities is essential.

The Cauchy-Riemann equations are $\boxed{u_x=v_y,u_y=-v_x}$. They give the gradient identityAt a regular intersection, where $f^{\prime }(z)\ne 0$, the two nonzero gradients are normal vectors to the two level sets, so their tangent directions are perpendicular. This is the orthogonality of regular real and imaginary level curves.

The regularity qualification is necessary: $|\nabla u|=|\nabla v|=|f^{\prime }|$, so at a critical point both normals vanish and the curves need not have unique tangents. For example $f=z^2$ gives diagonal lines for $u=0$ and coordinate axes for $v=0$, all meeting at the origin. Thus the conclusion is $\boxed{\text{right-angle intersection wherever }{f}^{\prime }(z)\ne 0}$, rather than an unrestricted claim at critical points.

Write $g=u+iv$ and let its constant modulus be $C$. If $C=0$, then $g=0$. For $C>0$, differentiating $u^2+v^2=C^2$ with respect to $x$ and $y$, then applying the Cauchy-Riemann equations, givesThusThe coefficient matrix has determinant $-(u^2+v^2)=-C^2\ne 0$, forcing $u_x=v_x=0$. The Cauchy-Riemann equations then also give $u_y=v_y=0$. Therefore $u,v$ are constant on a connected domain andThis proves the constant-modulus holomorphic function result. If a domain is disconnected, the conclusion applies separately to each connected component.

For an analytic function, the complex differentiability limit is independent of the direction of approach. A real increment gives $f^{\prime }(z)=u_x+i{v}_x$, whereas an imaginary increment givesEquality of real and imaginary parts proves the Cauchy-Riemann equations:Conversely, if $u,v$ have continuous first partial derivatives near a point and obey these equations, their real differentiability gives, for a complex increment $h$,Dividing by $h$ proves the complex differentiability there. Continuous first partial derivatives are a sufficient regularity hypothesis for this converse; agreement along the two coordinate directions alone is only a necessary condition.

Using the complex cosine identity,we findIts real part is exactly the prescribed function. Thus a harmonic conjugate is $v=y\cos x\cosh y-x\sin x\sinh y$, and the general answer isTo check the only freedom, two analytic functions with this real part differ by a function with zero real part. The Cauchy-Riemann equations then force both partial derivatives of its imaginary part to vanish, so it is constant on the connected domain. No normalization was supplied to fix $C$.

A function is an analytic function at $z_0$ if it is holomorphic throughout some open neighbourhood of $z_0$, equivalently if it has a convergent power series there. In real components $g=u+iv$, continuous first partial derivatives in a neighbourhood together with the Cauchy-Riemann equationsthroughout that neighbourhood are sufficient, and for an analytic function they hold there. Satisfying these equations only at the single point is not enough to establish analyticity in a neighbourhood.

Write $f=u+iv$ and $z=x+iy$. For a positively oriented, piecewise smooth simple closed contour bounding a region where $u,v$ have continuous first partial derivatives, Green's theorem givesThe Cauchy-Riemann equations are $u_x=v_y$ and $u_y=-v_x$, so both integrands vanish:This proves Cauchy's integral theorem under the standard regularity assumptions for Green's theorem. More general admissible contours follow by subdivision; every enclosed region must be free of singularities.

For a holomorphic function, the derivative supplied by complex differentiability must have the same value along real and imaginary increments. Those two computations giveEquating real and imaginary parts proves the Cauchy-Riemann equations:Conversely, if the first partial derivatives are continuous near a point and obey these equations there, real differentiability giveswhich proves complex differentiability under this regularity assumption.

The gradients of the real and imaginary parts have dot productand $|\nabla u{|}^2=|\nabla v{|}^2=|f^{\prime }(z){|}^2$. At every point with $f^{\prime }(z)\ne 0$, the two level sets are regular curves, their gradients are nonzero normal vectors, and their tangent directions are therefore perpendicular.

The printed phrase "at every point" needs this regular-point qualification. At a critical point with $f^{\prime }(z)=0$, the level sets may cross or fail to have a unique tangent. For example, $f(z)=z^2$ gives $u=x^2-y^2$ and $v=2xy$: at the origin the zero level sets are respectively the diagonals and the coordinate axes. Thus the valid conclusion is $\boxed{\text{orthogonality at every regular intersection, where }{f}^{\prime }(z)\ne 0}$.

The Cauchy-Riemann equations determine the harmonic conjugate. Firstso $v_y=u_x$ integrates with respect to $y$ toNowand the second Cauchy-Riemann equation, $v_x=-u_y$, forces $h^{\prime }(x)=0$. HenceThe normalization at zero gives $C=0$, so the required entire function is

At $z=x+iy$, compute the complex derivative first along real increments and then along imaginary increments. Holomorphy givesEquating real and imaginary parts proves the Cauchy-Riemann equations:The partial derivatives exist because these directional difference quotients are limits of the holomorphic derivative. This argument proves the necessary equations; it does not claim that their pointwise existence alone is sufficient for holomorphy.

Use the principal complex logarithm:The excluded nonpositive real ray allows a continuous choice of argument, and $L(1)=0$. For $z=x+iy$, its real part is $u=\frac{1}{2}\log (x^2+y^2)$. Differentiating the chosen local argument givesThese continuous partial derivatives satisfy the Cauchy-Riemann equations, proving holomorphy on the slit domain. Its derivative is

At every $z=x+iy\in U$, the Cauchy-Riemann equations areTo derive them from complex differentiability, approach $z$ along real increments $h$. The complex derivative givesApproaching along imaginary increments $ih$ instead givesThese limits exist because $f$ is holomorphic. Equating their real and imaginary parts proves the equations, together with $f^{\prime }=u_x+i{v}_x=v_y-i{u}_y$.