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
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.