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Let $f$ be holomorphic on the unit open disk, and continuous on the closure of that disk, such that $\forall \theta \in \mathbb{R}$ the value $f(e^{i \theta})$ lives on the boundary of the triangle with vertices $0,1,i$. What can I say about the values that $f$ takes and lives on the triangle? For example , there is some $|z_0|<1$ , such that $f(z_0)= \frac{1}{2} (1+i) $ or $f(z_0) = \frac{1}{10} (1+i) $? Maybe I can use Rouche, and the characterization of the elements of the convex hull, written as a linear combination. But I don't know how

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Non-constant holomorphic functions are continuous and open, so the boundary of the image of a domain is contained in the image of the boundary. In this case, the image of the open domain is bounded by a subset of (the boundary of) the triangle. Since the image is bounded, its boundary has to be the whole (boundary of the) triangle, so the image of the open unit disk is the open interior of the triangle.

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With that I can't conclude that these values are in the image set, but in the book, they ask me if it's true, can we say more ? –  Daniel Oct 30 '12 at 0:05
    
And how can I prove that the image of the boundary of the circle , is not surjective in the boundary of the triangle? –  Daniel Nov 3 '12 at 14:36
    
I don't quite understand the question, can you rephrase it? –  Lukas Geyer Nov 4 '12 at 22:26
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