Firstly, I don't know complex analysis.
Can the Laplace operator in a complex plane $z=x+iy$ be expressed as $$\nabla=\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}$$ or $$\nabla=\frac{\partial}{\partial x}+\frac{\partial}{i\partial y}=\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}$$ Which is right? or not one?
I originally wanted to verify the power series $V=C_n(x+iy)^n$ in complex plane $z=x+iy$ is exactly a solution for a Laplacian equation $$\nabla^2V=0$$. How to do that?
You can identify $\mathbb{R}^2 \cong \mathbb{C}$ by thinking of $z=x+iy$ with real numbers $x,y$.
Then, the following relationship holds:
If a function is holomorphic, then the functions $Re(f(x,y))=Re(f(z))$ and $Im(f(z))=Im(f(x,y))$ are harmonic as a function from $\mathbb{R}^2 \to \mathbb{R}$.
There is also the following relation, using Wirtinger derivatives (see https://en.wikipedia.org/wiki/Wirtinger_derivatives ):
$$ \frac{1}{4} \Delta u=\frac{\partial^2 u}{\partial z \partial \bar{z}} $$
Now, to look at your function we have that and using that $i^2=-1$: $$ \frac{\partial u}{\partial z}=\frac{1}{2}C_n(n(x+iy)^{n-1}-in(x+iy)^{n-1}i)= \\ C_nn(x+iy)^{n-1} $$ $$ \frac{\partial^2 u}{\partial z \partial \bar{z}}=\frac{1}{2}C_n ( n(n-1)(x+iy)^{n-2}+in(n-1)(x+iy)^{n-2}i)=0 $$ if your n=1, then the second step is not needed. If $n \geq 2$, the calculation is fine. If that was not the answer you were looking for, dont hesitate to ask!