I've managed to solve a dozen problems like these before, but I'm stuck at those including an unknown function such as the one below.
Determine all analytic functions $f(z)=u+iv$ with the real part of the form $u(x,y)=\varphi(x)(1-y)$ where $u(x,y)\in\mathcal{C}^2$ and which moreover satisfy $f(0)=0$. The answer should be expressed as a function of the variable $z=x+iy$.
What I've tried this far is of course to use the Cauchy-Riemann equations. They give us $$ v_x=-u_y=\varphi(x) \\ v_y=u_x=\varphi'(x)(1-y)=\varphi'(x)-y\varphi'(x). $$ Thus $$ v(x,y)=\int\varphi(x)dx+\Psi(y) \\ v(x,y)=y\varphi'(x)-\frac{y^2}{2}\varphi'(x)+\Phi(x). $$ This is where I get stuck. What I see is a crazy mixture of derivatives and integrals. The final result is $f(z)=a\left(z+\frac{iz^2}{2}\right)$, $a\in\mathbb{R}$, which obviously doens't contain $\varphi(x)$. Any help is appreciated.
$u(x,y)$ is a harmonic function, which means $u_{xx} + u_{yy} = 0$. Since $u_{yy} = 0$, this part is easy to solve (obtaining $u$ depending on two parameters, of which one is eliminated by the $f(0)=0$). Then use Cauchy-Riemann to get $v$.