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Take the punctured complex planeandIn polar coordinates, $u(r,\theta )=\log r$, soThus $u$ is harmonic on $D$.

If $u=\mathrm{Re}f$ for a holomorphic function $f$, the calculation from part (a) would giveThe integral of a derivative around a closed curve is zero, whereas the residue theorem givesThis contradiction proves that no such $f$ exists. It is precisely the log modulus has no global harmonic conjugate on the punctured plane obstruction.

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.