Let $f:\mathbb{C} \longrightarrow \mathbb{C}$ be a holomorphic function such that $\exists a,b:\mathbb{R} \longrightarrow \mathbb{C}$ such that $f(x+iy)=a(x)+ib(y)$ for every $(x,y) \in \mathbb{R}^2$. Prove that $\exists \lambda,k \in \mathbb{C}$ such that $f(z)=\lambda z+k$ .
I tried to use the condition given by the Wirtinger derivatives: $$\frac{\partial f}{\partial z}:=\frac{1}{2}\left(\frac{\partial f}{\partial x}-i \frac{\partial f}{\partial y}\right)=f'(z) \qquad \qquad \frac{\partial f}{\partial \bar{z}}:=\frac{1}{2}\left(\frac{\partial f}{\partial x}+i \frac{\partial f}{\partial y}\right)=0 $$ And the fact that $$ \frac{\partial f}{\partial x}=\frac{\partial a}{\partial x}+i\frac{\partial b}{\partial x} $$ and the same for $\partial y$, but this only gave me $f'(z)=2a'(x)$, which seems not going anywhere.
The answer really has been given by @Kenny Wong and others, but I'll spell it out a bit more:
Because $f$ is holomorphic the Cauchy-Riemann equations are satisfied: $$ a_x = b_y, \quad a_y = -b_x = 0 $$ Since $b_y$ is not a function of $x$ it follows that $a_x$ is not a function of $x$. Similarly as $a_x$ is not a function of $y$ it follows $b_y$ is not a function of $y$. Hence, there are constants $c,d,f \in \mathbb R$ such that \begin{align} a_x &= c \Rightarrow a(x) = cx + d \\ b_y &= c \Rightarrow b(y) = cy + f \end{align} So $$ f(z) = a(x) + ib(y) = cz + k, \quad k = d+if $$