Let $f$ be a holomorphic function at the unit circular disk and that $f(0)=0$. Now let $f$ be real valued function on the radii $[0,1)$ and $[0,e^{i\frac{\pi}{4}})$ Show that $f$ has a zero of, at least, degree $4$ in $z = 0$
I have no clue where to start!
Thanks in advance!
While @Robert answer suggests how to do the problem using the Taylor series and a local approach, let me show how to do it using the identity principle:
let $f_1(z)=\overline {f(\bar z)}-f(z)$ which is analytic in the disc;
since $f_1(r)=0, 0 \le r <1$ it follows $f_1(z)=0, z \in \mathbb D$ so $\overline {f(\bar z)}=f(z)$
and in particular $f(re^{-i\pi/4})=\overline {f(re^{i\pi/4})}=f(re^{i\pi/4}), 0 \le r <1$
Let $f_2(z)=f(iz)-f(z)$
Since $f_2(re^{-i\pi/4})=f(ire^{-i\pi/4})-f(re^{-i\pi/4})=f(re^{i\pi/4})-f(re^{-i\pi/4})=0, 0 \le r<1$ it follows as before $f_2(z)=0$ in the disc or $f(iz)=f(z)$
Conjugating (or using $f_3(z)=f(-iz)-f(z)$ zero for $z=re^{i\pi/4}$) we get $f(-iz)=f(z)$ so in particular $f(iz)=f(-iz)$ hence using $w=iz$ we get $f(w)=f(-w)$ too.
But now the symmetries above mean that $f(z)=\frac{f(z)+f(-z)+f(iz)+f(-iz)}{4}=g(z^4)$ for some analytic $g$ in the unit disc
(if $f(z)=\sum a_kz^k$ then $g(z)=\sum a_{4k}z^k$)
This immediately implies that if $f$ has a zero at $0$ it has a zero of order $4k, k \ge 1$ so the order is at least $4$ and more generally the next value possible is $8$ etc