Calculating a harmonic conjugate

Solution 1:

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 $$ v = \operatorname{Im}(F(z)). $$

Solution 2:

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 $f$.