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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?

julio_es_sui_glace
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Elise9
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    $au+bv = c$ means that all values $f(z)$ are on a line in $\Bbb C$. Now use https://math.stackexchange.com/q/181311/42969 – Martin R Sep 09 '23 at 17:03
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    You need to assume that $D$ is connected (at least). – copper.hat Sep 09 '23 at 17:24
  • second hint, in line with Steve's: use CR to sub out all instances of any partials of v. Then attempt to get something in the form of $(a^2 + b^2)______ = 0$. Conclude that _____ is 0, a similar argument will show that f' is 0, which is enough. – emesupap Sep 09 '23 at 17:41

4 Answers4

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

Steven Gubkin
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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.

copper.hat
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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.

Nitin Uniyal
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  • I combine $(3)$ and $(4)$ and I got $(a+b)u_x + (b-a)v_x = 0$. How did you obtain $u_x=v_x=0$? Btw, if $u_x = u_y = 0 $ that means $u(x,y)$ is constant. – Elise9 Sep 10 '23 at 09:08
  • @Elise9...See that the determinant of coefficient matrix of $u_x$ and $v_x$ is $a^2+b^2(\ne 0)$ which implies that there is only zero solution viz. $u_x=v_x=0$. This in turn suggests $u=const$. – Nitin Uniyal Sep 10 '23 at 16:04
  • @Elise9..Let me know if you still feel difficulty in the understanding of my answer. – Nitin Uniyal Sep 11 '23 at 03:04
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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.

user8675309
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