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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.

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up vote 4 down vote accepted

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 $v$?

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Thanks for the answer. I've done something different. Would be please take a look to it: Given that $a^2+b^2\neq0$ $\forall$ $(x,y)\in D$; Without loss of generality let $b\neq 0$ $\Rightarrow v(x,y)= {c-au(x,y)\over b}\Rightarrow$ $f(x,y)=u(x,y)+i[{c\over b}-{a\over b}u(x,y)]$ $\Rightarrow$ the images of D under $f$ lies on the straight line $y={c\over b}-{a\over b}x$ $\Rightarrow$ arg $f(z)$ is constant on D $\Rightarrow$ f is constant on D. – Sugata Adhya Dec 12 '12 at 15:11
@SugataAdhya: That's a neat argument! Looks good to me. – froggie Dec 12 '12 at 16:00
Thanks so much. – Sugata Adhya Dec 12 '12 at 16:59

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.

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  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)

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I think I've done something similar to your approach in the above comment. – Sugata Adhya Dec 12 '12 at 15:16
@SugataAdhya Yes, that works! (Even better when you solved it yourself!) – AD. Dec 12 '12 at 15:27
Thanks. Actually it occurred to me after I made the post. – Sugata Adhya Dec 12 '12 at 17:00

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