Corresponding analytic function?

Question

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?

Answer 1

Observe first that $$ z^3=(x+iy)^3=(x^3-3xy^2)+i(3x^2y-y^3), $$ and then conclude that $f(z)=(a+id)z^3+K$.

Answer 2

Given $v=3ax^2y-3dxy^2+ay^3+dx^3+K$

The corresponding analytic function is find out by "MILNE'S THOMSON METHOD"according to which

$$f(z)=\int\phi_1(z,0)dz+i\int\phi_2(z,0)dz+K$$

where $\phi_1(x,y)=(v_y)_{(x,y)}$ & $\phi_2(x,y)=(v_x)_{(x,y)}$

hence $\phi_1(z,0)=(v_y)_{(z,0)}$ & $\phi_2(z,0)=(v_x)_{(z,0)}$

Using above method you can easily find $v$.If you have any more doubt leave a coment

THANKS

Answer 3

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.