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a) Let $D$ be a domain whose boundary $C$ contains a straight-line segment $L$. Let $f(z)$ be analytic in $D$ and continous on $L$. Assume also that $\Im(f) = v(x,y)$ vanishes on $L$. Prove that $f$ is analytic on $L$.

I am a bit confused with what it means to be analytic on a line. I know how to show analyticity in a domain, via Morera, but for a line, I am not so sure.

b) Show that there is no function $f(z)$ analytic for $y>0$ and continuous for $y\geq0$ such that $$ f(z) = |x| $$

Thanks

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For a general subset $A$ of the plane, we say that a function is analytic on $A$ if it is analytic on some open set containing $A$. Should that be $f(z)=|z|$ in part (b)? –  Cameron Buie Apr 25 '13 at 20:50
    
The phrasing of the b part seems off. Perhaps the intention is to show there is no such function with $f(x)=|x|$ for real $x$? –  Harald Hanche-Olsen Apr 25 '13 at 20:57
    
Phrasing is correct. We have that $x$ is the real part of z –  Eager Student Apr 25 '13 at 21:37
    
If the phrasing is correct, it is definitely off. Saying there is no function with some properties such that … and following it with an explicit definition of the function, is just a very convoluted way of saying that the given function does not have the stated properties. –  Harald Hanche-Olsen Apr 25 '13 at 21:44
    
Hi Eager Student, just wanted to let you know that you can accept answers to all of your questions to signal to the person who answered that you are satisfied with their answer (among other reasons). Hope to see you around more! –  Antonio Vargas Apr 26 '13 at 16:27

2 Answers 2

Hint for part a: The Schwarz reflection principle, combined with a bit of rotation and translation.

Hint for part b, assuming the requirement is $f(x)=|x|$ for real $x$: Find a function that satisfies this around the positive real axis. Repeat with the negative axis. Can you reconcile the two?

Addition to part b: By reflection, you get an entire function $g$ so that $g(x)=|x|$ when $x$ is real. In particular $g(x)=x$ for $x>0$, hence by the principle of analytic continuation (the uniqueness part thereof) $g(z)=z$ for all $z$. But that clearly contradicts $g(x)=-x$ when $x<0$.

… okay, so that was overkill: Clearly, analyticity fails at the origin, and you don't need to invoke analytic continuation. But the argument as I gave it will work even if the origin is excluded from consideration.

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I see how after we translate and rotate to the real axis and use the reflection principle with the analytic continuation $$ g(w) = \left\{\begin{array}[lr] f(w) & Im(w)\geq 0 \\ \bar{f}(\bar{w}) & Im(w)\leq 0 \end{array}\right. $$ Thanks. For part b is it the same sort of thing. We have that $f$ is real on the real axis and analytic in the upper half plane so for $g$ defined above $g$ is entire and $f$ is constant, so it does not satisfy the condition. –  Eager Student Apr 27 '13 at 22:22
    
I think you need more for part b. I expanded my answer. –  Harald Hanche-Olsen Apr 28 '13 at 9:13

For part (b), I think it is a simple consequence of $C-R$ equations that a real valued analytic function defined in a domain $D$, is constant.

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