# Prove Function is Analytic

• Mar 29th 2008, 06:13 PM
ThePerfectHacker
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.
• Mar 30th 2008, 01:16 AM
Opalg
Quote:

Originally Posted by ThePerfectHacker
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. $w = \frac{z-i}{z+i}$.
• Mar 30th 2008, 09:23 AM
ThePerfectHacker
Quote:

Originally Posted by Opalg
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.

This is Mine 91:):)th Post!!!
• Mar 30th 2008, 12:40 PM
Opalg
Quote:

Originally Posted by ThePerfectHacker
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.