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
On
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.
On
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 $$ \frac{\partial^2 }{\partial z\partial \bar{z}}\left[ F(z) + G(\bar{z})\right] = \frac{\partial }{\partial \bar{z}}F'(z) + \frac{\partial }{\partial z} G'(\bar{z}) = 0 $$
On
The shortes proof of this is by splitting the operator: $$ (\delta^2_x + \delta^2_y)\phi = 0 $$ yields $$ ((\delta_x + i\delta_y)(\delta_x - i\delta_y))\phi = (\delta_x + i\delta_y)((\delta_x - i\delta_y)\phi) = (\delta_x + i\delta_y)f = 0 $$ or, as well $$ ((\delta_x - i\delta_y)(\delta_x + i\delta_y))\phi = (\delta_x - i\delta_y)((\delta_x + i\delta_y)\phi) = (\delta_x - i\delta_y)g = 0 $$ We must now solve the two transport equations $$ (\delta_x + i\delta_y)f = 0 $$ $$ (\delta_x - i\delta_y)g = 0 $$ Their differential operators are complex scaling-rotators, which means that the transport equations are a left-sense and a right-sense scaling-rotation. We have thus advantage to rewrite the transport equations in polar coordinates $ (r,\phi) $: $$ (\delta_x + i\delta_y)f = 0 \quad becomes \quad (\delta_\phi + i r \delta_r)F = 0 $$ which is solved by any twice differentiatable function $$ F(re^{-i\phi}) $$ and $$ (\delta_x - i\delta_y)g = 0 \quad becomes \quad (\delta_\phi - i r \delta_r)G = 0 $$ which is solved by any twice differentiatable function $$ G(re^{i\phi}) $$ The general solution of the Laplace equation is thus $$ \Phi = F(re^{-i\phi}) + G(re^{i\phi}) $$ i.e. a superposition of forward- and backward-rotating waves; or, in Cartesian coordinates $$ \phi = f(x + iy) + g(x-iy) $$
QED
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].