Let function $f=u +iv$ is analytic on $D$ and for some $a,b,c \in \mathbb R$, $a^2+b^2 \neq 0$ and $au+bv=c$ on $D$. Prove that $f$ is constant on $D$

Question

Let function $f=u +iv$ be analytic on some domain $D$. Let $a,b,c \in \mathbb R$ such that $a^2+b^2 \neq 0$ and $au+bv=c$ on $D$. Prove that $f$ is constant on $D$.

What I have done is following: Actually I only thought that Can I use Liouville's Theorem because we already have $f$ is analytic and if it is bounded that would imply that $f$ is constant by Liouville's theorem. But how can I show that $f$ is bounded?

After seeing answer I write these: We have $au+bv=c$ and C-R equations which are $u_x = u_y$ and $u_y = - v_x$ then we get \begin{align} \label{eq1} au_x +b v_x &= 0 \tag{1} \\ \label{eq2} au_y + b v_y &= 0 \tag{2} \end{align} If I substitute $u_y$ and $v_y$ with appropiate pairs in \eqref{eq2} then $$\label{eq3} au_x +b v_x = 0 \tag{3}$$ and $$\label{eq4} -av_x + b u_x = 0\tag{4}$$ Combining these two equations \eqref{eq3},\eqref{eq4}, $$(a+b)u_x + (b-a)v_x = 0 \tag{5}$$

And I stop here. Could help a little more?

Answer 1

Hint: differentiate both sides of $au + bv = c$ with respect to $x$ and $y$. Use the Cauchy-Riemann equations.

Answer 2

Here is a proof assuming that $D$ is open & connected. Note that $\lim_{t \to 0} {f(z+th)-f(z) \over t} = f'(z)h$. Hence $a \operatorname{re} (f'(z)h ) + b \operatorname{im} (f'(z)h ) = 0$ for all $h$. If $a \neq 0$ choose $h=\overline{f'(z)}$ to get $f'(z) = 0$, and similarly if $a=0$. Hence $f$ is constant.

Answer 3

With the given condition $a^2+b^2\ne 0$, solve $(3)$ and $(4)$ simultaneously to get $u_x=v_x=0$ or equivalently $u_x=0$ and $u_y=0$ (By CR-equations). Now what does it tell about $u(x,y)$? Use Liouville's theorem then.

Answer 4

notice that $a-ib\neq 0$
$g:=(a-ib)\cdot f -c=i\cdot(av -bu)$ is analytic and purely imaginary hence constant on each connected component. Justification: either open mapping theorem, or max modulus theorem applied to $\exp\circ g$ and examining its derivative. Conclude $f =\frac{g+c}{a-ib}$ is constant as well.