Laplace's equation, theoretical doubt I tried to solve Laplace equation in two dimensions
$$u_{xx} + u_{yy}=0$$
without any condition added. Just tried to solve it by separation of variables, so I assumed that the solution be of the form $u(x,y) = \Phi_x(x) \cdot \Phi_y(y)$.
Replacing the correct derivatives in the equation, I obtained
$$\Phi''_x(x) \cdot \Phi_y(y) + \Phi_x(x) \cdot \Phi''_y(y) =0 $$
and separating the functions in the two independent variables, yields:
$$\Phi''_x(x) \cdot \Phi_y(y) = -\Phi_x(x) \cdot \Phi''_y(y)  \quad\implies\quad \frac{\Phi''_x(x)}{\Phi_x(x)}=-\frac{\Phi''_y(y)}{\Phi_y(y)} $$
If this equality is true, then: 
$$
\begin{cases}
\frac{\Phi''_x(x)}{\Phi_x(x)}=K^2\\
-\frac{\Phi''_y(y)}{\Phi_y(y)}=K^2
\end{cases}
$$
Solving the two ODES, I got:
$$
\begin{cases}
\Phi_x(x) = a_1 e^{Kx} + a_2 e^{-Kx} \\
\Phi_y(y) = b_1 e^{iKy} + b_2 e^{-iKy}
\end{cases}
$$  
So, putting in order these computations:
$$ u(x,y) = \Phi_x(x) \cdot \Phi_y(y) = a_1 b_1 e^{K(x+iy)} + a_1 b_2 e^{K(x-iy)}+ a_2 b_1 e^{K(-x+iy)} + a_2 b_2 e^{K(-x-iy)}$$
This solution surely verifies the Laplace equation (by substitution). 
But I observed that exponential functions are expressed in function of
$$x+iy, \quad -x+iy, \quad x-iy, \quad -x-iy$$
We certainly note that $-x+iy$ can be expressed in function of $x-iy$ up to a constant factor. 
We also note that $-x-iy$ can be expressed in function of $x+iy$ up to another constant factor. 
Therefore, the solution becomes expressable like 
$$u(x,y) = f_1(x+iy) + f_2(x-iy)$$
samely valid solution and easily verifiable by subsitution.
My doubt is the following: is my way of reasoning correct? Are my solutions suitable? 
I just started to solve my first PDEs, so I am a full novice and I am not used to express the solutions in power series nor using other methods.
Is my way to solve it well-posed or is there any theoretical weakness? Thanks in advance. Forgive my naiveness.

You have the Laplace equation 
$$\partial_{x}^{2}\Psi(x, y)+\partial_{y}^{2}\Psi(x, y)=0$$
By letting
$$\xi=x+iy$$
$$\bar{\xi}=x-iy$$
The equation becomes
$$\partial_{\xi\bar{\xi}}^{2}\Psi(\xi, \bar{\xi})=0$$
Integrating with respect to $\xi$ we get
$$\partial_{\bar{\xi}}\Psi(\bar{\xi}, \xi)=f(\bar{\xi})$$
Integrating with respect to $\bar{\xi}$
$$\Psi(\bar{\xi}, \xi)=\int{f(\bar{\xi})}d\bar{\xi}+G(\xi)=F(\bar{\xi})+G(\xi)$$
Thus the general solution is of the form
$$\Psi(x, y)=F(x-iy)+G(x+iy)$$
