# finding the degree of a zero of a function with given conditions

Mathematics Asked by pencil321 on January 3, 2022

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 $$[0,e^{ifrac{pi}{4}})$$ Show that $$f$$ has a zero of, at least, degree $$4$$ in $$z = 0$$

I have no clue where to start!

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 $$f(re^{-ipi/4})=overline {f(re^{ipi/4})}=f(re^{ipi/4}), 0 le r <1$$

Let $$f_2(z)=f(iz)-f(z)$$

Since $$f_2(re^{-ipi/4})=f(ire^{-ipi/4})-f(re^{-ipi/4})=f(re^{ipi/4})-f(re^{-ipi/4})=0, 0 le r<1$$ 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 $$z=re^{ipi/4}$$) 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

Hint: suppose the zero at $$z=0$$ has degree $$k$$. What does the Maclaurin series of $$f(z)$$ look like? What does that say about $$f(z)$$ when $$z$$ is close to $$0$$ and is on one of those two radii?