3
$\begingroup$

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.

3 Answers 3

4

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 $v$?

  • 0
    @SugataAdhya: That's a neat argument! Looks good to me.2012-12-12
1

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.

1

Hint:

  1. An analytic function with constant real (or imaginary) part, must be constant (why?).
  2. 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)

  • 0
    Thanks. Actually it occurred to me after I made the post.2012-12-12