To show that $f$ is constant on $D$ Let $f:D$ (domain) $\rightarrow \mathbb C:(x,y)\mapsto u(x,y)+ iv(x,y)$ be analytic on $D$ & $\exists$ $a, b, c \in \mathbb R$ such that (i) $a^2+b^2\neq0$ & (ii) $au(x,y)+bv(x,y)=c$ $\forall$ $(x,y)\in D$. I need to show that $f$ is constant on $D$. A clue rather than a detailed solution would be appreciated. Thanks.
 A: Hint: Take the equation $au + bv = c$ and take the partial derivative of both sides first with respect to $x$, then with respect to $y$. You then get two equations. Using these two equations and the Cauchy-Riemann equations, what can you conclude about $u$ and $v$?
A: The Cauchy Riemann equations must be satsified:
\begin{gather}\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\\
\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}
\end{gather}
Differentiating $au(x,y)+bv(x,y)=c$ with respect to $x$ and then $y$ gives
\begin{gather}\frac{\partial u}{\partial x}=\frac{-b}{a}\frac{\partial v}{\partial x}\\
\frac{\partial u}{\partial y}=\frac{-b}{a}\frac{\partial v}{\partial y}
\end{gather}
Therefore,
$$\frac{\partial u}{\partial x}=\frac{-b}{a}\frac{\partial v}{\partial x}=\frac{b}{a}\frac{\partial u}{\partial y}=\frac{b}{a}\frac{-b}{a}\frac{\partial v}{\partial y}=-\frac{b^2}{a^2}\frac{\partial u}{\partial x}
$$
This of course implies $\frac{\partial u}{\partial x}=0$. I think the OP can do the rest.
A: Hint: 


*

*An analytic function with constant real (or imaginary) part, must be constant (why?).

*Now what can we say about the function $$z=x+iy\mapsto (a-ib)f(z)$$


(Note that the condition on $a,b$ is just that they are not both zero)
