1
$\begingroup$

Let $f$ be analytic in a domain (open connected set of the domain of definition) (that is, every point of that domain is such that $f'$ exists in that point and also in a neighborhood of radius $r>0$ of that point). Show that his absolute value $|f|$ can't be constant unless $f$ is constant too.

Firstly, I'm a little confused about the statement. I must prove $|f|$ is constant $\Rightarrow f$ is constant?

Assuming that's what I must show, here's my attempt: if $f$ and $\overline{f}$ are both analytic then $f$ is constant so if $f$ is not constant and $f$ is analytic we have that $\overline{f}$ is not analytic at some point $p$. But $|f| = \sqrt{f\overline{f}}$ so $|f|$ is also not analytic at the point $p$ so it can't be constant because constant functions are analytics at all points of their domain.

Is my attempt correct? Thanks in advance.

$\endgroup$
  • 1
    $\begingroup$ I mean, one direction is trivial. If $f$ is constant then certainly taking its absolute value will not change anything and $|f|$ is still constant. For the other direction, what theorems do you know? The maximum principle for example? Liouville's theorem? $\endgroup$ – Luke Oct 2 '17 at 12:30
  • $\begingroup$ I know what I used in my attempt and also that if $f=u+iv$ is analytic then $ -v+iu$ is analytic too $\endgroup$ – AnalyticHarmony Oct 2 '17 at 13:11
  • 1
    $\begingroup$ How do you get from the non-analyticity of $\overline{f}$ to that of $\sqrt{f\overline{f}}$? Different approach: If $\lvert f\rvert \equiv 0$, it's trivial that $f \equiv 0$. If $\lvert f\rvert \neq 0$, look at the analytic [assuming $\lvert f\rvert$ constant] function $\frac{\lvert f\rvert^2}{f}$. $\endgroup$ – Daniel Fischer Oct 2 '17 at 13:22
3
$\begingroup$

Suppose that $|f|\equiv k$ for some $k\in[0,+\infty)$.

  1. If $k=0$, then $f\equiv0$.
  2. Otherwise, the image of $f$ is contained in the circle centered at $0$ with radius $k$. Such a set contains no non-empty subset of $\mathbb C$. But, if $f$ was not constant, then by the open mapping theorem, its image would be an open (and obviously non-empty) subset of $\mathbb C$.
$\endgroup$
  • $\begingroup$ Thanks! Could you check my proof please? $\endgroup$ – AnalyticHarmony Oct 2 '17 at 13:38
  • 1
    $\begingroup$ @NoGoodAtMath It looks good to me, although some passages need more details. I hope you don't mind that I added the proof-verification tag to it. $\endgroup$ – José Carlos Santos Oct 2 '17 at 13:43
3
$\begingroup$

Observe that $|f|^2=u^2+v^2$, where $f(x,y)=u(x,y)+iv(x,y)$. If $|f(z)|^2=0$ for some $z \in \Omega$, we are done. Therefore consider $|f|^2 \neq 0$. Differeniate $|f|^2$ with respect to $x$ and $y$. Since $|f|^2$ is constant, we get by Cauchy-Riemann that $$ 0=2uu_x+2vv_x=2uu_x-2vu_y $$ $$ 0=2uu_y+2vv_y=2 u u_y+2vu_x $$ Rewriting it in matrix form with the matrix $M$ gives: $$ 2 \begin{bmatrix} u & -v \\ v & u \end{bmatrix} \begin{bmatrix} u_x \\ u_y \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \end{bmatrix} $$ Then, observe that $det(M)=|f|^2\neq0$. There, in order for the linear equation system to be statisfied, it is required that $u_x=u_y=0$. Therefore, $Re(f)$ is constant. Therefore $f$ is constant.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.