# Corresponding analytic function?

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?

• Bug alert! It should be $v=3ax^2y-3dxy^2\color{red}-ay^3+dx^3+K$ Feb 22, 2018 at 5:25

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.

$$u + iv = a x^3 - 3dx^2 y - 3axy^2 + dy^3 + i(3ax^2y - 3dxy^2 + ay^3 + dx^3 + K)$$

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.

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

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