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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,

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    $\begingroup$ Holomorphic functions of constant modulus are constant $\endgroup$ Commented Nov 20, 2015 at 0:24
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    $\begingroup$ 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 $\mathbb{C}$? $\endgroup$ Commented Nov 20, 2015 at 0:50
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    $\begingroup$ 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). $\endgroup$ Commented Nov 20, 2015 at 0:56
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    $\begingroup$ "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 $\mathbb D.$ $\endgroup$
    – zhw.
    Commented Nov 20, 2015 at 1:02
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    $\begingroup$ 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 $\endgroup$
    – zhw.
    Commented Nov 20, 2015 at 1:14

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