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I'm struggling a bit to arrive at the conclusion that $f(z)$ is a constant.

Suppose $f(z)$ is holomorphic on and inside the unit disk, and that it has no zeroes on the interior. Also, assume that $|f(e^{i\theta}|$ = 1, for $0<\theta< 2\pi$.

Then by the maximum principle, the maximum of $f(z)$ is attained on the boundary -- that is, the unit circle. But since $g:=\large \frac {1}{f(z)}$ is holomorphic on and inside the unit disk, too, then it also attains its maximum on the boundary, and so this forces $f(z)$ to be minimal on the circle as well.

Knowing now that $f(z)$ attains both its max and min on the circle, how can I conclude that it is constant? This is tricky of course since $f(z)$ could assume infinitely many different values that have modulus = 1.

So, I hope to make use of the open mapping theorem.

Is this too ambitious? If I can say that the open set, namely the interior of the unit disk, maps onto the unit circle in the w-plane, then we know that the image is compact, and by Heine-Borel's theorem, the image is closed and bounded. In particular, the image is closed. By the open mapping theorem, the holomorphic function $f(z)$ must be constant.

...unfortunately I don't see why $f(z)$ has to be an onto mapping carrying the interior of the unit disk onto the unit circle in the w-plane.

Any hints or comments are welcome.

Thanks,

User001
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    Holomorphic functions of constant modulus are constant – Henricus V. Nov 20 '15 at 00:24
  • The interior must map to the unit circle because $f(z)$ has modulus 1 everywhere in it. It's sort of what it means to map to the unit circle. – Arturo don Juan Nov 20 '15 at 00:27
  • Hi @ArturodonJuan, is it necessarily an onto mapping? If not, and the image is merely an arc of the circle, then the image may not be closed ... – User001 Nov 20 '15 at 00:29
  • Thanks for the cool link @HenryW -- it's nice to see the CR-equations come up again; I would not have thought of going for a f'(z) = 0 claim, so I will study what's on that link soon. For now, I feel that I am close enough with the open mapping theorem, and will try a bit more in this direction to see whether I can make it work ... – User001 Nov 20 '15 at 00:31
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    Why must the function be onto in order for the image to be compact? More precisely, how isn't any arc/point, or collection of them, on the unit circle compact with respect to $\mathbb{C}$? – Arturo don Juan Nov 20 '15 at 00:50
  • Right @ArturodonJuan, I was just thinking of this, too. Hmm...I am struggling with this line of reasoning, though, since I feel that a subset of the circle would be something analogous to an open interval on the real line...and hence not a closed set...where am I going wrong with this logic? Perhaps I need to think of an easy sequential argument to make regarding convergent sequences converging to a point inside of the subset / arc ... – User001 Nov 20 '15 at 00:52
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    The topology of $\mathbb{C}$ coincides with the topology on $\mathbb{R}^2$ (in the obvious way). With that said, any finite line (e.g. arc of a circle) or finite collection of points in $\mathbb{R}^2$ is immediately compact, as for every open cover for $\mathbb{R}^2$ there is certainly a finite open subcover that contains that finite line and/or collection of points. This can intuitively be seen as coming from the fact that open in $\mathbb{R}^2$ is defined in 2-dimensions (i.e. every open set must have a non-zero area). – Arturo don Juan Nov 20 '15 at 00:56
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    "On and inside the unit disc" doesn't make much sense. A better assumption is that $f$ is continuous on $\overline {\mathbb D}$ and holomorphic in $\mathbb D.$ – zhw. Nov 20 '15 at 01:02
  • Hi @ArturodonJuan, ... an open set in 2-dimensions must have non-zero area ... but a line, an arc, has no area and hence is a closed set in $R^2$ - and equivalently, in $C$. Is this correct? Thanks so much, – User001 Nov 20 '15 at 01:06
  • Hi @zhw, yes, the assumption was analyticity on and inside the unit disk, but there is always possibility of transcription error. I am reading 1997 exam questions, transcribed by former students, I believe. Also, IIRC analyticity on the circle is not needed ... – User001 Nov 20 '15 at 01:09
  • Hi @ArturodonJuan, thanks so much for your time and for helping me with my conceptual mistake of what it means for a set to be open in one dimension, compared with the meaning of open set in two-dimensions. Have a great night :-) – User001 Nov 20 '15 at 01:13
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    Since such transcriptions can be in error, you should think about them before posting a question. It's important that you understand why "on and inside the unit disc" is strange language – zhw. Nov 20 '15 at 01:14
  • Hi @zhw., hmm...is there an obvious triviality with the assumption of analyticity on the circle? – User001 Nov 20 '15 at 01:19
  • Or perhaps you are referring to issues with the definition of differentiability on the boundary @zhw., and that we usually assume differentiability on the interior of a set ... – User001 Nov 20 '15 at 01:22
  • Btw, @zhw., I posted a very cool harmonic function question last night, if you'd like to look at it :-). I tried it for a very long time, but have not gotten anywhere with it. For now, I have to move on, but feel free to check it out :-) – User001 Nov 20 '15 at 01:23

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