a) For which values of $a,b,c,d$ is the function $u(x,y)=ax^3+bx^2y+cxy^2+dy^3$ harmonic?
b) Find a harmonic conjugate $v(x,y)$ of $u(x,y)$ in this case.
c) Find an analytic function $f(z)=u(x,y)+iv(x,y)$, where $z=x+iy$.
a) Use the Laplace equation for $u$ to get $$(\partial_x^2+\partial_y^2)u(x,y)=(6a+2c)x+(2b+6d)y=0,$$ thus $c=-3a,b=-3d$.
b) $$u(x,y)=ax^3-3dx^2y-3axy^2+dy^3.$$ Harmonic conjugate satisfies the Cauchy-Riemann condition $$v_y=u_x=3a(x^2-y^2)-6dxy,\\ v_x=-u_y=3d(x^2-y^2)+6axy.$$ Integrating these, we have $$v(x,y)=dx^3+3ax^2y-3dxy^2-ay^3.$$ Alternatively, we can put $$f(z)=2u(\dfrac{z}{2},\dfrac{z}{2i})-u(0,0).$$ Then $$\begin{align}f(z)&=2\bigg(a(\dfrac{z}{2})^3-3d(\dfrac{z}{2})^2(\dfrac{z}{2i})-3a(\dfrac{z}{2})(\dfrac{z}{2i})^2+d(\dfrac{z}{2i})^3\bigg)\\&=(a+3id+3a+id)\dfrac{z^3}{4}\\&=(a+id)z^3\\&=(a+id)(x+iy)^3\end{align}$$ The imaginary part is $$v(x,y)=dx^3+3ax^2y-3dxy^2-ay^3.$$ c) $$\begin{align}f(z)&=u(x,y)+iv(x,y)\\&=ax^3-3dx^2y-3axy^2+dy^3+i(dx^3+3ax^2y-3dxy^2-ay^3)\\&=a(x^3+3ix^2y-3xy^2-iy^3)+d(ix^3-3x^2y-3ixy^2+y^3)\\&=a(x+iy)^3+id(x+iy)^3\\&=(a+id)z^3\end{align}$$ Remark: The coefficients $1,3,3,1$ hint the use of $(?+?)^3$. We can also substitute $y=0,x=z$ to get $(a+id)z^3$ immediately!
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