General solution to Laplace Equation Show the general solution to the Laplace equation, $$\frac{\partial^2\phi}{\partial x^2}+\frac{\partial ^2\phi}{\partial y^2}=0$$
is $\phi(x,y)=f(x+iy)+g(x-iy)$.
The only thought I have is let $x+iy$ be $z$, a complex number so $\phi=f(z)+g(z^*)$. What are the next steps
 A: You are asked to show that the general solution to an equation has a particular form, so you should start from the equation. You spotted correctly that it is a good idea to make a change of variables, $$z=x+iy, \qquad \bar{z} = x-iy.$$ By the Chain Rule, you can verify that $$\frac{\partial}{\partial x} = \frac{\partial}{\partial z} + \frac{\partial}{\partial \bar{z}} \quad \text{and} \quad \frac{\partial}{\partial y} = i\left( \frac{\partial}{\partial z} - \frac{\partial}{\partial \bar{z}} \right).$$ Let me also write $\phi(x,y) = \psi(z, \bar{z})$ for the new function that we want to solve for after making the change of variables. It is an easy calculation to check that $$\frac{\partial^2}{\partial x^2} \phi =\left(\frac{\partial}{\partial z} + \frac{\partial}{\partial \bar{z}} \right) \left( \frac{\partial}{\partial z} + \frac{\partial}{\partial \bar{z}} \right) \psi = \frac{\partial^2\psi}{\partial z^2} + 2 \frac{\partial^2 \psi}{\partial z \partial \bar{z}} + \frac{\partial^2 \psi}{\partial \bar{z}^2},$$ and that $$\frac{\partial^2}{\partial y^2} \phi =(i)^2 \left(\frac{\partial}{\partial z} - \frac{\partial}{\partial \bar{z}} \right) \left( \frac{\partial}{\partial z} - \frac{\partial}{\partial \bar{z}} \right) \psi = -\frac{\partial^2\psi}{\partial z^2} + 2 \frac{\partial^2 \psi}{\partial z \partial \bar{z}} - \frac{\partial^2 \psi}{\partial \bar{z}^2}.$$ So your equation becomes $$4 \frac{\partial^2 \psi}{\partial z \partial \bar{z}} = 0.$$ Can you see how to finish it off? [Hint: integrate].
A: Let's enforce the substitution $z=x+iy$ and $\bar z=x-iy$.  Then, we have
$$\begin{align}
\frac{\partial \phi}{\partial x}&=\frac{\partial \phi}{\partial z}\frac{\partial z}{\partial x}+\frac{\partial \phi}{\partial \bar z}\frac{\partial \bar z}{\partial x}\\\\
&=\frac{\partial \phi}{\partial z}+\frac{\partial \phi}{\partial \bar z}
\end{align}$$
and 
$$\frac{\partial^2 \phi}{\partial x^2}=\frac{\partial^2 \phi}{\partial z^2}+2\frac{\partial^2 \phi}{\partial z \partial \bar z}+\frac{\partial^2 \phi}{\partial \bar z^2} \tag 1$$
Similarly, for the partial derivative with respect to $y$, we have
$$\frac{\partial^2 \phi}{\partial y^2}=-\frac{\partial^2 \phi}{\partial z^2}+2\frac{\partial^2 \phi}{\partial z \partial \bar z}-\frac{\partial^2 \phi}{\partial \bar z^2} \tag 2$$
Adding $(1)$ and $(2)$, we obtain
$$\frac{\partial^2 \phi}{\partial x^2}+\frac{\partial^2 \phi}{\partial y^2}=4\frac{\partial^2 \phi}{\partial z \partial \bar z}=0 \tag 3$$
whereupon solving $(3)$ for $\phi$ reveals that $\phi(z,\bar z) = f(z)+g(\bar z)$.  Finally, substituting back for $z$ and $\bar z$ yields
$$\phi(x,y)=f(x+iy)+g(x-iy)$$
as was to be shown.
A: If we look at
$$
\partial_{xx}\phi + \partial_{yy}\phi = \left(\partial_x + i\partial_y\right)\left(\partial_x - i\partial_y\right)\phi = 0
$$
a solution can be any that takes the form of $z = x+iy$ or $\bar{z} = x-iy$ can be used to transform the latter expression as
$$
\frac{\partial^2 \phi}{\partial z\partial \bar{z}} = 0
$$
which implies a general solution
$$
\phi = F(z) + G(\bar{z})
$$
so taking that
$$
\frac{\partial^2 }{\partial z\partial \bar{z}}\left[ F(z) + G(\bar{z})\right] = \frac{\partial }{\partial \bar{z}}F'(z) + \frac{\partial }{\partial z} G'(\bar{z}) = 0
$$
