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Thread: Prove Function is Analytic

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    Prove Function is Analytic

    Let f: \{z\in \mathbb{C}| \Im (z) > 0 \} \mapsto \mathbb{C} be continous on \{ z\in \mathbb{C}|\Im (z) > 0\} and analytic on \{ z\in \mathbb{C}|\Im (z) > 0 \} \setminus \{ z\in \mathbb{C}: |z| = 1\} prove that f is analytic on \{ z\in \mathbb{C} | \Im(z) > 0\}.

    My idea was to transform this problem into a simpler one. If f is analytic on a region minus a line segment and continous there, then it must be analytic everywhere on the region. Thus, perhaps a conformal map will transform the circle into a line segment.
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    Quote Originally Posted by ThePerfectHacker View Post
    Let f: \{z\in \mathbb{C}| \Im (z) > 0 \} \mapsto \mathbb{C} be continous on \{ z\in \mathbb{C}|\Im (z) > 0\} and analytic on \{ z\in \mathbb{C}|\Im (z) > 0 \} \setminus \{ z\in \mathbb{C}: |z| = 1\} prove that f is analytic on \{ z\in \mathbb{C} | \Im(z) > 0\}.

    My idea was to transform this problem into a simpler one. If f is analytic on a region minus a line segment and continous there, then it must be analytic everywhere on the region. Thus, perhaps a conformal map will transform the circle into a line segment.
    You can certainly find a conformal map that will transform the upper half-plane to the unit disc and take the unit circle to a line, e.g. w = \frac{z-i}{z+i}.
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    Quote Originally Posted by Opalg View Post
    You can certainly find a conformal map that will transform the upper half-plane to the unit disc and take the unit circle to a line, e.g. w = \frac{z-i}{z+i}.
    Do you agree then that this completes the proof? Because f\left( \frac{i+iz}{1-z} \right) is analytic on unit disk because it is continous the line segment (using the theorem). And so f is analytic on upper-half plane.

    Note: My professor said that any function can be analytically continued along an analytic curve. Meaning if f is analytic on a region minus an analytic curve, but continous there. Then in fact f is analytic on the curve too.

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    Quote Originally Posted by ThePerfectHacker View Post
    Do you agree then that this completes the proof? Because f\left( \frac{i+iz}{1-z} \right) is analytic on unit disk because it is continous the line segment (using the theorem). And so f is analytic on upper-half plane.

    Note: My professor said that any function can be analytically continued along an analytic curve. Meaning if f is analytic on a region minus an analytic curve, but continous there. Then in fact f is analytic on the curve too.
    I don't know that theorem, but it sounds very plausible. Your prof sounds as though he knows what he's talking about.
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