# f analytic on unit disk implies f is real on (-1,1)?

Looking at a problem in Wunsch. Suppose f is analytic on the unit disk. Then I am asked to show that f must be real on the interval (-1,1). If f = u + i v, then how do I show that v(x,0) = 0 for $-1<x<1$ ?

Secondly, I then have to show that every derivative of $f$ is also real.

-
You need more conditions, $f$ could be a constant. –  J. J. May 1 '13 at 17:03
Indeed, that is where this is heading. If I can show the above, then I will be able to show that $f(\overline{z})=\overline{f(z)}$ on the unit disk. Then if I am given that $f(z)$ and $f(\overline{z})$ are both analytic on the unit disk, then $f(z)$ will be constant. But I need a reason as to why $v(x,0)=0$ for $-1<x<1$, even if $f$ is constant. –  David May 1 '13 at 17:20
Looks like I have a counterexample. $f=u(x,y)+iv(x,y)=-2xy+i(x^2-y^2)$. Using Cauchy-Riemann conditions, the function $f$ is analytic on the unit disk, but it is not real on the interval $(-1,1)$. For example, $f(0.5,0)=-2(0.5)(0)+i(0.5^2-0^2)=0.25i$. –  David May 1 '13 at 20:40
It's quite clear that either the textbook problem is incomplete, or your reading of it is. –  75064 May 1 '13 at 20:45
Turns out there was an error in the statement of the problem in the textbook. The errata page adds that the function must be real on the real axis. –  David May 1 '13 at 21:49