In Ahlfors textbook (p.72 of the 3rd edition) is the proof of the statement:

An analytic function $f$ in a region $\Omega$ whose argument $arg(f)$ is constant must be a constant function.

First he proves that if $Re(f)$ is a constant function then $f$ is constant. I've understood his proof for that. And now I quote his proof for the original statement:

"... if $arg(f)$ is constant, we can set $u=kv$ $\space$with constant $k$ (unless $v$ is identically zero). But $u-kv$ is the real part of $(1+ik)f$, and we conclude again that $f$ must be reduce to a constant."

I can't see why this shows that $f$ is constant, I would appreciate if someone could explain me this proof.

  • $\begingroup$ What are $u$ and $v$? $\endgroup$ – Michael Albanese Jan 3 '14 at 12:44
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    $\begingroup$ In general (at least in the textbooks I've read), if $f:\mathbb C \to \mathbb C$ and $z=x+iy$, $f(z)=u(x,y)+iv(x,y)$. $\endgroup$ – user100106 Jan 3 '14 at 13:00
  • $\begingroup$ Choosing to call the real and imaginary parts $u$ and $v$ respectively is completely arbitrary. Even though this naming convention is common, you should still specify what $u$ and $v$ are. $\endgroup$ – Michael Albanese Jan 3 '14 at 13:08
  • $\begingroup$ arg(f) being constant also contradicts the open mapping theorem. $\endgroup$ – Steven Gubkin Jan 3 '14 at 16:00
  • $\begingroup$ Could you explain that? I didn't understand the comment. $\endgroup$ – user100106 Jan 3 '14 at 16:03

But $u - kv$ is the real part of $(1+ik)f$

and $u-kv$ is constant [unless $v \equiv 0$, when we observe that $f$ is real-valued, hence constant]. So by the fact proved before, $(1+ik)f$ is constant. But since $1+ik$ is a nonzero constant, that is equivalent to the constancy of $f$.

  • $\begingroup$ One last question, I think that what I don't understand of the proof is: why if $arg(f)$ is constant $\implies$ $u$ can we written as $u=kv$ ($k$ constant)? (sorry, I didn't even get what my doubt was, now I've realized it). I've understood everything else from the proof. $\endgroup$ – user100106 Jan 3 '14 at 12:56
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    $\begingroup$ $\arg f$ being constant means the image of $f$ is contained in one ray emanating from $0$, in particular, in a straight line through $0$. Every straight line through $0$ other than the real axis can be defined by an equation $x = k\cdot y$, and that means you have $u = kv$ for the real resp. imaginary parts of $f$; unless the range of $f$ is contained in the real line, but then we are (modulo a multiplication with $i$) in the situation treated before. $\endgroup$ – Daniel Fischer Jan 3 '14 at 13:01
  • $\begingroup$ Thanks!, I finally got it. $\endgroup$ – user100106 Jan 3 '14 at 13:02
  • $\begingroup$ In this question, is $arg(f)$ an element of the domain, i.e the argument of $f(z)$ is $z$; or is it argument as in Arg $z = \theta$ where $-\pi < \theta < \pi$? $\endgroup$ – Al Jebr Mar 28 '16 at 7:04
  • $\begingroup$ @AlJebr The latter, $f(z) = \lvert f(z)\rvert \cdot \exp \bigl(i\arg f(z)\bigr)$, though it's not necessary to assume it's in $(-\pi,\pi)$. $\endgroup$ – Daniel Fischer Mar 28 '16 at 14:25

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