I have the following equation $$\frac{\partial^2 V}{\partial x^2 }+\frac{\partial^2 V}{\partial y^2}=-4\pi(x^2+y^2)$$ Such that $V(x,y)$ is real and $V(x,y)=0$ at $y=0$.
As it is clear form the problem the characteristic equation has complex roots. I have following as the solution $$z=\phi_1(y+ix)+\phi_2(y-ix)-\dfrac{\pi x^2\left(6\left(y-\mathrm{i}x\right)^2+\left(3\mathrm{i}-3\right)x^2+\left(8\mathrm{i}+4\right)x\left(y-\mathrm{i}x\right)\right)}{3}$$
$z(x,y)$ is the sum of the solution of the associated homogenous PDE and any particular solution.
I cannot see how you get a so complicated particular solution : $$z_p=-\dfrac{\pi x^2\left(6\left(y-\mathrm{i}x\right)^2+\left(3\mathrm{i}-3\right)x^2+\left(8\mathrm{i}+4\right)x\left(y-\mathrm{i}x\right)\right)}{3}$$ Moreover this function seems not correct. Did you put it into the PDE to check if the equation is satisfied ?
Without having the details of your calculus it is not possible to find where a possible mistake occurred.
There is a much simpler way to find a particular solution. For example, look for a particular solution on the form $$z_p=f(x)+g(y)$$ $$f''(x)+g''(y)=-4\pi(x^2+y^2)$$ $$\begin{cases} f''(x)=-4\pi(x^2) & ;\quad f(x)=-\frac{\pi}{3} x^4 \\ g''(y)=-4\pi(y^2) & ;\quad g(x)=-\frac{\pi}{3} y^4 \end{cases}$$ No need for more terms in $f(x)$ and in $g(y)$ since we are not looking for many particular solutions but only for one. Doesn't matter which one. $$z_p=-\frac{\pi}{3} (x^4+y^4)$$ $$z=\phi_1(y+ix)+\phi_2(y-ix)-\frac{\pi}{3} (x^4+y^4)$$ Of course, since any particular solution is convenient, other particular solutions could be used instead of the above one. So, the solution can be written on an infinity of equivalent forms. If you find a solution which at first sight looks different from the above, check if they are not equivalent , due to different form of functions $\phi_1$ and $\phi_2$ .