The following equation, $$(\partial_x + i\partial_y)u - c(\partial_x+i\partial_y)au=0$$
($a=a(x,y)$ and $\partial_x=\frac{\partial}{\partial x}$)
with solution, $$u=\exp(ca)f(x+iy)$$
where $f$ and $g$ are arbitrary entire functions, a is some scalar function and $c$ is a scalar.
How can I derive the solution ?
On
I would solve this via characteristics and a variant of integrating factors. (It took me a while to realize that your operator on the second term is the $\bar{\partial}$ derivative of $a$, not $au$.) Let's multiply our PDE by $\exp(-ca)$. Then our PDE becomes
$$0 = \exp(-ca)(\partial_x+i\partial_y)u-c\exp(-ca)(\partial_x+i\partial_y)a\cdot u \Longrightarrow 0 = (\partial_x+i\partial_y)(\exp(-ca)u).$$
Suppose $v = \exp(-ca)u$ is actually a function of another variable $t$: $v(x(t),y(t))$, then
$$\frac{d}{dt}v(x(t),y(t)) = \partial_x v\frac{dx}{dt}+\partial_y v\frac{dy}{dt}.$$
If we set $\frac{dx}{dt} = 1$ and $\frac{dy}{dt} = i$, we get that
$$\frac{d}{dt}v = 0$$
by our above PDE. For this solution set, $x = t+C$ and $y = it+D$. Further our above ODE says that $v$ is constant on our characteristic curve. There is some freedom in how we parameterize our characteristic curves (as seen by our constants $C$ and $D$) but if we reparameterize, we can in fact just take $x = t$ and $y= it+D$.
Since $\frac{dv}{dt} = 0$, we just need to determine its value at $0$ and we will have a general solution for $v$. Let $v(0) = f(y_0)$ (since $x_0 = x(0) = 0$). But then $y_0 = y(0) = D$. We also have that $ix=it$ and so we can solve for $D$: $D = y-ix$. Thus
$$v(t) = v(x(t),y(t)) = f(D) = f(y-ix) = f(-i(x+iy)) = g(x+iy),$$
where $g$ is some entire function (since $v$ solves $\bar{\partial} v = 0$). Now just assemble the pieces.
On
$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\pars{\partial_{x} + \ic\partial_{y}}{\rm u}\pars{x,y} -c\pars{\partial_{x} + \ic\partial_{y}}{\rm a}\pars{x,y}{\rm u}\pars{x,y} = 0:\ {\large ?}}$
Your notation is confused. For the second term, I'll assume $\ds{\pars{\partial_{x} + \ic\partial_{y}}}$ is 'applied' just to $\ds{\rm a}$. It means I understand $\quad\ds{\pars{\partial_{x} + \ic\partial_{y}}{\rm au}}\quad$ as $\ds{\bracks{\pars{\partial_{x} + \ic\partial_{y}}{\rm a}}{\rm u}}$ In that case, we can rewrite the equation as: $$ \fermi\pars{x,y}{\rm u}\pars{x,y}=0\,,\qquad\qquad \fermi\pars{x,y} \equiv \bracks{1 - \pars{\partial_{x} + \ic\partial_{y}}{\rm a}\pars{x,y}} $$ The solution will be $\ds{{\rm u}\pars{x,y} = 0}$ unless $\ds{\fermi\pars{x,y} = 0}$.
Hint: Your equation can be written
$$\partial_x (\log u-ca) + i\partial_y(\log u - ca) = 0$$
This equation has the solution $\log u - ca = g$ for any $g$ that satisfy
$$\partial_x g + i\partial_yg = 0$$