Is it correct to say that equation
$$ u_{xx} +u_{yy}=0 $$
has as solution $ u(x,y)=\Phi_1(y+ix) + \Phi_2(y-ix) $ ?
I am not intended to solve it with any condition. I just wanted to solve (via polynomial $\lambda^2 +1 =0$) the pure equation, regardless of physical meaning. Can solution written be left in this way or is there any other way to express it without involving $i$ constant (like in 2nd order ODEs)? thanks.
Your proposal is almost correct. In fact, this is a conclusion from complex analysis. The conclusion reads that any real-valued harmonic function (i.e., solution to the Laplace equation) must be the real part (or imaginary part, as per your preference) of a holomorphic function, and vise versa.
The way for its proof more or less follows your idea. Define \begin{align} z&=x+iy,\\ \bar{z}&=x-iy. \end{align} It is not hard to verify that $$ \Delta=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}=4\frac{\partial^2}{\partial z\partial\bar{z}}. $$ Therefore, $$ \Delta u=0\iff\frac{\partial^2u}{\partial z\partial\bar{z}}=0. $$ However, $$ \frac{\partial^2u}{\partial z\partial\bar{z}}=0\iff\frac{\partial}{\partial\bar{z}}\left(\frac{\partial u}{\partial z}\right)=0, $$ which implies that $\partial u/\partial z$ is holomorphic. Thus $$ \frac{\partial u}{\partial z}=g'(z) $$ for some holomorphic $g$. Integrate this last equation, and one obtains $$ u(z,\bar{z})=g(z)+\overline{h(z)}, $$ where $h$ is another holomorphic function.
Finally, since $u$ is real valued, it is a must that $h=g+K$ with $K\in\mathbb{R}$. Therefore, $$ u=\Re\left(2g+K\right), $$ meaning that $u$ is the real part of some holomorphic function.
Conversely, if $g$ is holomorphic, it is straightforward to check that its real part is a harmonic function: $$ \Delta\left(\Re\left(g\right)\right)=4\frac{\partial^2}{\partial z\partial\bar{z}}\frac{g(z)+\overline{g(z)}}{2}=0. $$
Back to the real expression. If $u$ is harmonic, then $$ u(x,y)=\Re\left(2g(z)\right)=g(z)+\overline{g(z)}=g(x+iy)+\bar{g}(x-iy) $$ for some holomorphic $g$. Here $\bar{g}$, as a different function from $g$, is defined as $$ \bar{g}(z)=\overline{g(\bar{z})}. $$