# Transform a complex function in (x,y) into one in z?

I was working on a problem that asked us to find the harmonic conjugate of $$u=x^3-3xy^2$$.

After determining that $$u$$ is indeed harmonic by using Laplace's equation $$\nabla u^2=\frac{\partial ^2}{\partial x^2}\left(u\right)+\frac{\partial ^2}{\partial y^2}\left(u\right)=0$$, I determined its harmonic conjugate to be $$v(x,y)=3x^2y-y^3+C$$ by using the Cauchy-Riemann equations $$u_x=v_y$$ and $$u_y=-v_x$$.

Now, the complex function is $$f(x,y)=u(x,y)+iv(x,y)=(x^3-3xy^2)+i(3x^2y-y^3)+iC$$.

I would like to transform this into $$f(z)$$ instead of $$f(x,y)$$. How would I go about doing this?

I have tried to substitute the equations $$x=\frac{z+\overline{z}}{2}$$ and $$y=\frac{z-\overline{z}}{2i}$$:

$$f(x,y)=(x^3-3xy^2)+i(3x^2y-y^3)+iC$$, which becomes:

$$f(z)=\left(\frac{z+\overline{z}}{2}\right)^3-3\left(\frac{z+\overline{z}}{2}\right)\left(\frac{z-\overline{z}}{2i}\right)^2+i\left[3\left(\frac{z+\overline{z}}{2}\right)^2\left(\frac{z-\overline{z}}{2i}\right)-\left(\frac{z-\overline{z}}{2i}\right)^3\right]+iC$$.

But this seems to simplify with a $$\overline{z}$$ in the numerator. It is my understanding that an analytic function will not have $$\overline{z}$$? I am not sure what I am doing wrong.

$$\displaystyle z^3 = (x + iy)^3 = x^3 + 3ix^2 y - 3xy^2 - iy^3.$$
$$\displaystyle f(x,y) = (x^3 - 3xy^2) + i(3x^2 y - y^3) + C = z^3 + C.$$