Analytic function with zero on |z|=1/2 If f is an analytic function defined on open unit disc s.t. f is zero on |z|=1/2. Does it necessarily implies that f is a constant?
I tried to apply the theorem which states that " An analytic function which is zero on a part of a line is identically zero". 
But, does this theorem implies that any function not fulfilling the hypothesis of being zero on a part of line is a non-constant?
Any hint will help me. Thanks in advance.
 A: Grubby, first principles proof:
An analytic function has a Taylor series expansion. Since $f$ is analytic on the open unit disc,
$$ f(z) = \sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!}z^k $$
converges for $\lvert z \rvert <1$. We also have
$$ f^{(k)}(0) = \frac{1}{2\pi i} \int_{|z|=1/2} \frac{f(z)}{z^{k+1}} \, dz, $$
using the Cauchy integral formula. But the numerator is precisely equal to zero, so the integral is zero, every coefficient of the Taylor series is zero, and hence $f \equiv 0$ on the whole disc.
Obviously this proof only works for this special case. Otherwise, you will have to use some variant of the identity theorem: probably the most useful is

Let $f$ be analytic on the domain $D$, and let $(a_n)$ be a sequence of distinct points with a limit point in $D$. If $f(a_n)=0$ for every $a_n$, then $f \equiv 0$ on $D$.

(You will recall this is proven using isolation of zeros, which is proven by using that zeros of $f$ have finite orders: at a zero $a$, $f(z)=(z-a)^k g(z)$, $g \neq 0$ on a small disc around $a$.)
If you apply isolation of zeros, you get exactly the result you want.
