Calculating a harmonic conjugate
Sebastian Wright 
Is the following reasoning correct?
Determine a harmonic conjugate to the function \begin{equation} f(x,y)=2y^{3}-6x^{2}y+4x^{2}-7xy-4y^{2}+3x+4y-4 \end{equation}
We first of all check if $f(x,y)$ is indeed a harmonic function. This amounts to show $f(x,y)$ satisfy the two-dimensional Laplace equation\begin{equation} \frac{\partial^{2 }f}{\partial x^{2}}+\frac{\partial^{2} f}{\partial y^{2}}=0 \tag{1} \end{equation}We have $\frac{\partial^{2}f}{\partial x^{2}}=8-12y$ and $\frac{\partial^{2} f}{\partial y^{2}}=12y-8$. Thus, (1) is fulfilled, and so $f(x,y)$ is harmonic.
Next, we seek to determine a harmonic conjugate to the given function. Let $u(x,y)=2y^{3}-6x^{2}y+4x^{2}-7xy-4y^{2}+3x+4y-4$. \begin{equation*} u_{x}=v_{y} \iff -12xy+8x-7y+3=v_{y} \end{equation*}Integrate with respect to $y$\begin{equation} v=-6xy^{2}+8xy-\frac{7}{2}y^{2}+3y+h(x) \tag{2} \end{equation}where $h(x)$ is a function of $x$ alone. To determine this, we use the second Cauchy-Riemann equation $v_{x}=-u_{y}$ \begin{align*} -u_{y}=v_{x} &\iff 6x^{2}+7x-6y^{2}+8y-4=h'(x)-6y^{2}+8y \\ &\iff h'(x)=6x^{2}+7x-4 \end{align*}Integrating with respect to $x$ we have\begin{equation} h(x)=2x^{3}+\frac{7}{2}x^{2}-4x+C \end{equation}where $C$ is an arbitrary constant. Therefore, if we let $C=0$, then one harmonic conjugate of $u$ is given as:\begin{equation} v=2x^{3}+\frac{7}{2}x^{2}-6xy^{2}+8xy-4x-\frac{7}{2}y^{2}+3y \end{equation}
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$\begingroup$Yet another shortcut. Since $u$ is harmonic (on the simply connected domain $\mathbb{C}$), there has to be a harmonic conjugate $v$. Let $F = u+iv$ be the corresponding holomorphic function. It follows from (the derivation of) Cauchy-Riemann's equations that: $$ F' = u'_x - i\,u'_y = -12xy + 8x -7y + 3 + i(6x^2+7x-6y^2+8y-4). $$ Let $G(z) = 3 + 8z + i(6z^2+7z-4)$. Then $G(z) = F'(z)$ if $z$ is real, so by the identity theorem, $G = F'$ for all $z$. Hence $$ F(z) = 3z + 4z^2 - 4 + i(2z^3+\frac72z^2-4z+C) $$ for some real constant $C$ (the real part of the constant of integration has to be $4$ to match $u$). Finally $$ v = \operatorname{Im}(F(z)). $$
$\endgroup$ 1 $\begingroup$Your reasoning is correct. I'll give an alternative solution, which works only for polynomials, but can be carried out mechanically: substitute $x=(z+\bar z)/2$ and $y=(z-\bar z)/(2i)$. Result: $$ f(z) = iz^3+ (2+7i/4)z^2+(3/2-2i)z -i\bar z^3 + (2-7i/4) \bar z^2 + (3/2+2i)\bar z - 4$$ The reason this polynomial is harmonic is that no monomial has $z$ and $\bar z$ together. The fact that we started with something real lends it certain symmetry: namely, the terms $$-i\bar z^3 + (2-7i/4) \bar z^2 + (3/2+2i)\bar z$$ are just the complex conjugates of $iz^3+ (2+7i/4)z^2+(3/2-2i)z$. Therefore, $$ f(z) = \operatorname{Re}(2iz^3+ 2(2+7i/4)z^2+2(3/2-2i)z -4)$$ and the imaginary part of same complex polynomial gives the harmonic conjugate of $f$.
$\endgroup$ $\begingroup$why does f(x,y) need to be harmonic? Can someone explain? An analytic function is strongly connected with harmonic function, but if it's not harmonic is it then not analytical? Bit confused by the termonology here..
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