Codex @codex  0

Differentiate the prescribed imaginary part with respect to $y$:The Cauchy-Riemann equation $u_x=v_y$ integrates toIts $y$ derivative is $u_y=-e^x[(x+1)\sin y+y\cos y]+F^{\prime }(y)$. On the other hand, $-v_x=-e^x[(x+1)\sin y+y\cos y]$. The second Cauchy-Riemann equation therefore requires $F^{\prime }(y)=0$. Combining the real and imaginary parts,Thus the most general answer on a connected domain isA real constant preserves the given imaginary part. Conversely, the difference of any two such analytic functions has zero imaginary part, and its Cauchy-Riemann derivatives force it to be constant. This proves completeness, rather than only exhibiting one possible harmonic conjugate.

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.

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 $C$.

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

Set $A=x^2-y^2$ and $B=2xy$. Direct differentiation givesTheir sum vanishes, proving that $u$ is a harmonic function. More generally the Cauchy-Riemann equations give $u_{xx}+u_{yy}=v_{yx}-v_{xy}=0$ and $v_{xx}+v_{yy}=-u_{yx}+u_{xy}=0$; the mixed partial derivatives commute because a holomorphic function is smooth. Since $e^{i{z}^2}=e^{-B}(\cos A+i\sin A)$, a harmonic conjugate and the resulting holomorphic function areThese expressions also verify the Cauchy-Riemann equations directly.

Set $A=x^2-y^2$ and $B=2xy$. Direct differentiation givesTheir sum vanishes, proving that $u$ is a harmonic function. More generally the Cauchy-Riemann equations give $u_{xx}+u_{yy}=v_{yx}-v_{xy}=0$ and $v_{xx}+v_{yy}=-u_{yx}+u_{xy}=0$; the mixed partial derivatives commute because a holomorphic function is smooth. Since $e^{i{z}^2}=e^{-B}(\cos A+i\sin A)$, a harmonic conjugate and the resulting holomorphic function areThese expressions also verify the Cauchy-Riemann equations directly.

A meromorphic function on the unit disc has bounded characteristic if and only if it has this representation. Its pole Blaschke product $B$ makes $g=Bf$ analytic with bounded positive logarithmic means. The least harmonic majorant by expanding disk lifts gives $H\ge {\log }^{+}|g|$. A harmonic conjugate gives an analytic $A$ with $\mathrm{Re}A=H$, so $u=g{e}^{-A}$ and $v=B{e}^{-A}$ are bounded. Conversely the Nevanlinna first main theorem bounds the characteristic of $u/v$ by that of $u$ and $1/v$, hence by a constant.

For real boundary data $u$, the displayed holomorphic integral has real part equal to the disk Poisson integral of $u$ and imaginary part its normalized harmonic conjugate. If $u\ge 0$ is not identically zero, the real part is strictly positive inside the disk. A holomorphic logarithm then has imaginary part between $-\pi /2$ and $\pi /2$, while its real part records the logarithm of the magnitude. This separates uniform imaginary-part control from large real parts on small boundary sets.

For a positive harmonic function on the unit disc, choose a global harmonic conjugate and apply the Carathéodory derivative bound for the right half-plane to $u+iv$. This gives $|\nabla u(0)|\le 2u(0)$ by the Cauchy-Riemann equations. A disk automorphism sending zero to $z$ has derivative modulus $1-|z{|}^2$, so the chain rule gives the displayed bound. Composing a boundary-pole Poisson kernel with a disk automorphism attains equality, proving the coefficient is optimal at every point.