I have found a general harmonic function of form $a x^3 – 3dx^2 y – 3axy^2 + dy^3$ and it's harmonic conjugate $v = 3ax^2y – 3dxy^2 + ay^3 + dx^3 + K$ where k is constant. I now am asked to find the corresponding analytic function $f(z) $ expressed in terms of $z$, and to check up to an imaginary constant $f(z) = 2u(\frac{1}{2}z, \frac{1}{2i}z) – u(0,0)$.
I know that an function $f(z)$ is analytic if its derivative is continuous at $z$, and it should be of form (I imagine) $u(x,y) + iv(x,y)$, but not how to find such a function. If I could have some pointers that would be great.
Update: Substituting $f(z) = u(x,y) + iv(x,y)$ I have managed to find $f(z) = z^3(a + di)$ but I still don't understand the last part?
Best Answer
The really clever trick to find $f(z)$ when we know that $f = u+iv$ and we have formulas for $u(x,y)$ and $v(x,y)$ is the following: Put $f = u + iv$, substitute $y=0$ and $x=z$ and your expession for $f(z)$ magically appears.
For your example:
$$ u + iv = a x^3 - 3dx^2 y - 3axy^2 + dy^3 + i(3ax^2y - 3dxy^2 + ay^3 + dx^3 + K)$$
Put $y=0$ and $x = z$ to obtain:
$$f(z) = az^3 + idz^3 + iK = (a+id)z^3 + iK.$$
This method works, but only if you already know that $u$ and $v$ are conjugate harmonic functions.
(An alternative, but much more cumbersome metod is to put $x = \frac12(z + \bar z)$ and $y = \frac{1}{2i}(z-\bar z)$, expand and simplify.)
Why does the trick work?
Let $g(z) = (a+id)z^3 + iK$. By construction $g(z) = f(z)$ whenever $z$ is real. Hence, by the identity theorem for holomorphic functions $g(z) = f(z)$ everywhere.