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A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 3.8

Suppose $f$ is holomorphic in region $G$, and $f(G) \subseteq \{ |z|=1 \}$. Prove $f$ is constant.

I will now attempt to elaborate the following proof at a Winter 2017 course in Oregon State University.

What errors if any are there? Or are there more elegant ways to approach this? I have a feeling this can be answered with Ch2 only, i.e. Cauchy-Riemann or differentiation/holomorphic properties instead of having to use Möbius transformations.

OSU Pf: Let $g(z)=\frac{1+z}{1-z}$, and define $h(z)=g(f(z)), z \in G \setminus \{z : f(z) = 1\}$. Then $h$ is holomorphic on its domain, and $h$ is imaginary valued by Exer 3.7. By a variation of Exer 2.19, $h$ is constant. QED


My Pf: $\because f(G) \subseteq C[0,1]$, let's consider the Möbius transformation in the preceding Exer 3.7 $g: \mathbb C \setminus \{z = 1\} \to \mathbb C$ s.t. $g(z) := \frac{1+z}{1-z}$:

If we plug in $C[0,1] \setminus \{1\}$ in $g$, then we'll get the imaginary axis by Exer 3.7. Precisely: $$g(\{e^{it}\}_{t \in \mathbb R \setminus \{0\}}) = \{is\}_{s \in \mathbb R}. \tag{1}$$ Now, define $G' := G \setminus \{z \in G | f(z) = 1 \}$ and $h: G' \to \mathbb C$ s.t. $h := g \circ f$ s.t. $h(z) = \frac{1+f(z)}{1-f(z)}$. If we plug in $G'$ in $h$, then we'll get the imaginary axis. Precisely: $$h(G') := \frac{1+f(G')}{1-f(G')} \stackrel{(1)}{=} \{is\}_{s \in \mathbb R}. \tag{2}$$

Now Exer 2.19 says that a real valued holomorphic function over a region is constant: $f(z)=u(z) \implies u_x=0=u_y \implies f'=0$ to conclude by Thm 2.17 that $f$ is constant or simply by partial integration that $u$ is constant. Actually, an imaginary valued holomorphic function over a region is constant too: $f(z)=iv(z) \implies v_x=0=v_y \implies f'=0$ again by Cauchy-Riemann Thm 2.13 to conclude by Thm 2.17 that $f$ is constant or simply by partial integration that $v$ is constant.

$(2)$ precisely says that $h$ is imaginary valued over $G'$. $\therefore,$ if $G'$ is a region (A) and if $h$ is holomorphic on $G'$ (B), then $h$ is constant on $G'$ with value I'll denote $Hi, H \in \mathbb R$:

$\forall z \in G',$

$$Hi = \frac{1+f(z)}{1-f(z)} \implies f(z) = \frac{Hi-1}{Hi+1}, \tag{3}$$

where $Hi+1 \ne 0 \forall H \in \mathbb R$.

$\therefore, f$ is constant on $G'$ (Q4) with value given in $(3)$.

QED except possibly for (C)


(A) $G'$ is a region

I guess if $G \setminus G'$ is finite, then G' is a region. I'm thinking $D[0,1]$ is a region and then $D[0,1] \setminus \{0\}$ is still a region.

(B) To show $h$ is holomorphic in $G'$:

Well $h(z)$ is differentiable $\forall z \in G'$ and $f(z) \ne 1 \forall z \in G'$ and $f'(z)$ exists in $G' \subseteq G$ because $f$ is differentiable in $G$ because $f$ is holomorphic in $G$.

$$h'(z) = g'(f(z)) f'(z) = \frac{2}{(1-w)^2}|_{w=f(z)} f'(z) = \frac{2 f'(z)}{(1-f(z))^2} $$

Now, $f'(z)$ exists on an open disc $D[z,r_z] \ \forall z \in G$ where $r_z$ denotes the radius of the open disc s.t. $f(z)$ is holomorphic at $z$. So, I guess $\frac{2 f'(z)}{(1-f(z))^2} = h'(z)$ exists on an open disc with the same radius $D[z,r_z] \ \forall z \in G'$, and $\therefore, h$ is holomorphic in $G'$.


(C) Possible flaw:

It seems that on $G'$, $f$ has value $\frac{Hi-1}{Hi+1}$ while on $G \setminus G'$, $f$ has value $1$.

$$\therefore, \forall z \in G, f(z) = \frac{Hi-1}{Hi+1} 1_{G'}(z) + 1_{G \setminus G'}(z)$$

It seems then that we've actually show only that $f$ is constant on $G$ except for the subset of G where $f=1$.

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    $\begingroup$ This is all very complicated: "to show $h$ is holomorphic in $G'$?". Why not just appeal to the fact that the composite of two holomorphic functions is holomorphic? $\endgroup$ – Lord Shark the Unknown Jul 28 '18 at 9:18
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    $\begingroup$ If it's not, then so much the worse for that text. $\endgroup$ – Lord Shark the Unknown Jul 28 '18 at 9:35
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    $\begingroup$ I'd use the open mapping theorem.... A more naive way is to apply Cauchy-Riemann. $\endgroup$ – Lord Shark the Unknown Jul 28 '18 at 9:43
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    $\begingroup$ What do you mean by $f(G) \subseteq \{ z=1 \}$? $\endgroup$ – zhw. Aug 5 '18 at 16:27
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    $\begingroup$ But you wrote $z=1$ not $|z|=1$ $\endgroup$ – zhw. Aug 5 '18 at 18:26
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The Cauchy-Riemann equations have a geometric interpretation. Let $f$ be holomorphic at $a$ and let $f'(a)\ne0$. Consider the horizontal line through $a$ consisting of points $a+s$ for real $s$, and also the vertical line through $a$, that is the points $a+it$ for $t$ real. Then these are mapped by $f$ into two curves $C_1$ and $C_2$ meeting at $f(a)$. Cauchy-Riemann implies these meet at right angles there.

But if the image of $f$ were within a 1-dimensional subspace such as the unit circle, then $C_1$ and $C_2$ would be restricted within too, which means they cannot intersect orthogonally. The only way out of this impasse is for $f'(a)=0$. This must happen for all $a$.

An introductory book that makes much of such geometric interpretations is Needham's Visual Complex Analysis (OUP).

If you really don't like geometry, write $f(x+iy)=u+iv$ in the usual way. If $f$ maps to the unit circle, then $u^2+v^2=1$. Differentiating gives $uu_x+vv_x=uu_y+vv_y=0$. Cauchy-Riemann gives $-uv_x+vu_x=0$. Then $$u_x=u^2u_x+v^2u_x=u(uu_x+vv_x)+v(-uv_x+vu_x)=0$$ and similarly $v_x=0$. Therefore $f'=0$.

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  • $\begingroup$ I just knew there had to be a way to do this with just Cauchy-Riemann. Suddenly, I no longer care about the kinda inelegant Möbius approach. Thanks! $\endgroup$ – BCLC Aug 6 '18 at 5:43
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Note that $f\overline {f}=1$ in $G.$ This implies $\overline {f} =1/f$ in $G.$ Hence $\overline {f}$ is holomorphic in $G.$ This implies both $f+\overline f= 2\text { Re } f$ and $f-\overline f=2i\text { Im } f$ are holomorphic in $G.$ By the remarks you made right after $(2)$ in your question, these functions are constant, which implies $f$ is constant.

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  • $\begingroup$ I just knew there had to be a way to do this with just regular differentiation/holomorphic stuff and not Möbius. Thanks! $\endgroup$ – BCLC Aug 6 '18 at 16:00
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    $\begingroup$ Thanks. I changed your edit as that page is not what I would refer a reader to. I was referring to the remarks in your question right after $(2);$ no need for anything else. $\endgroup$ – zhw. Aug 6 '18 at 16:28
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In every point on a line in the plane the possible tangent vectors form a real monodimensional vector space. An holomorphic function has in every point of its domain a derivative map that is a complex linear, that is a roto-homothetic real transformation of the plane.

Turning attention to your problem: in every point of the region $G$ the tangent vectors form a bidimensional real vector space. When a tangent vector is transformed by the derivative map it is rotated and/or enlarged in the plane of the tangent vectors in the image of the point. But the only possible tangent vectors in any point of the image (subset of the unit circle, that is a set of arcs and/or points) are in a real monodimensional vector space (for points on an arc) or zero-dimensional vector space (for isolated points). So for that roto-homothetic transformation to be fitted, it must make any tangent vector in the region $G$ vanish, that is, the derivative map must be $0$.

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  • $\begingroup$ HWAT. At this point, the reader is only 3 chapters into complex analysis and is not expected to have a topology background apart from basic topology in elementary analysis. Anyhoo, my understanding of your answer is...that that's your best (as in most elementary proof you can think of) guess for Q5, and you're not answering Q1-4? $\endgroup$ – BCLC Jul 28 '18 at 10:06
  • $\begingroup$ trying, edited question w/c is now reopened. $\endgroup$ – BCLC Aug 5 '18 at 14:06

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