I was trying to solve $$\frac{{\partial}^4\phi}{\partial{z^2}\partial\bar{z}^2}=0$$
Using the Wirtinger Derivatives $$\frac{\partial}{\partial z} = \frac{1}{2}(\frac{\partial }{\partial x} - i\frac{\partial }{\partial y})$$
$$\frac{\partial}{\partial \bar{z}} = \frac{1}{2}(\frac{\partial }{\partial x} + i\frac{\partial }{\partial y})$$
I used the following by multiplying the two wirtinger derivatives by itself, $$\frac{\partial^2}{\partial {z}^2} = \frac{1}{4}(\frac{\partial^2 }{\partial x^2} - 2i\frac{\partial^2 }{\partial y \partial x}-\frac{\partial^2 }{\partial y^2})$$
$$\frac{\partial^2}{\partial \bar{z}^2} = \frac{1}{4}(\frac{\partial^2 }{\partial x^2} + 2i\frac{\partial^2 }{\partial y \partial x}-\frac{\partial^2 }{\partial y^2})$$
And then I fit into the previous equation I wanted to solve and then got two linear partial differential equations by solving the real part and the imaginary part.
I'm having trouble how to solve them simultaneously.
Please help.
Well, the whole point of using Wirtinger derivatives is to make this kind of problem easy. One can "trivially" see that $\phi = (\overline{z}+a)p(z) + (z+b)q(\overline{z})$ is a solution, for arbitrary functions p and q. Why is this "trivial"? Because $z$ and $\overline{z}$ behave as if they were (linearly) independent variables. Posing the question you did is very similar to (the same as?) asking for solutions to $$\frac{\partial^4\phi}{\partial^2s\partial^2t}=0$$ for two linearly independent variables $s$ and $t$. (And oh, by the way, you'd get a mess if you tried to solve the above, using $s=u+v$ and $t=u-v$ - the same mess that you are currently getting).