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If $f(z)$ is analytic function and satisfies $ \vert f(z)\vert <1$ for $\vert z\vert<1$. show if $f(z)$ has a zero of order m at $z_o$ , then $\vert z_o \vert^ m > \vert f(0)\vert$.

here is what I know: if we let $\psi = (z-z_o)/ ( 1-\bar z_o z )$ then $\vert f(z) \vert < \vert \psi(z)\vert^m$ but how can I prove the last inequality? i.e. how to prove that $\vert f(z) \vert < \vert \psi (z) \vert^m$?

I just think that Im missing something? Im interested specifically in solving it by using this way $\psi$

thanks

Danny
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Consider $$ g(z)=\left(f\circ \psi^{-1}\right) (z),$$ where $$\psi(z)=\frac{z-z_0}{1-\bar{z_0}z},\quad\psi ^{-1}(z)=\frac{z+z_0}{1+\bar{z_0}z}.$$ Then $g$ satisfies $|g(z)|<1$ for $|z|<1$ and $g$ has a zero of order $m$ at $z=0$. So we have $$|g(z)|\le|z|^m.\tag{1}$$

Put $z=-z_0$ in $(1)$, we have $$ |-z_0|^m\ge|g(-z_0)|=|f\circ \psi ^{-1}(-z_0)|=|f(0)|,$$ which proves the inequality.

ts375_zk26
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  • can you please explain how did you get that $\vert g(z)\vert <1 $ ? Im doing it now and I dont know what is I`m doing wrong. – Danny Apr 04 '17 at 12:30
  • Write $\zeta =\psi ^{-1}(z)=\frac{z+z_0}{1+\bar{z_0}z}$. Then $$|\zeta |<1$$ for $z$ with $|z|<1,$ since $\psi(z)=\frac{z-z_0}{1-\bar{z_0}z}$ is a bijective mapping of the unit disk to the unit disk. Thus $$|g(z)|=|f\circ \psi ^{-1}(z)|=|f(\psi ^{-1}(z))|=|f(\zeta )|<1$$ for $(|z|<1)$. – ts375_zk26 Apr 04 '17 at 21:37
  • pardon me, but how did you get $\vert g(z)\vert < \vert z^m \vert$ – Danny Apr 05 '17 at 00:43
  • Sorry, $(1)$ should be $|g(z)\le |z|^m$ and the next expression should be $$|-z_0|^m\ge |g(-z_0)|=|f\circ \psi ^{-1}(-z_0)|=|f(0)|,$$ – ts375_zk26 Apr 05 '17 at 02:18
  • Since $g(z)=f\circ \psi ^{-1}(z)$ has a zero of order $m$ at $z=0$, $\frac{g(z)}{z^{m-1}}$ is analytic in ${|z|<1}$ and $g(0)=0$, $|g(z)|<1$. So, by Scwarz lemma, $$ \left|\frac{g(z)}{z^{m-1}}\right|\le |z|,$$ hence $|g(z)|\le |z|^m.$ – ts375_zk26 Apr 05 '17 at 03:24
  • We can not prove the strict inequality $$|z_0|^m>|f(0)|.$$ A counter-example: Let $f(z)=\left(\frac{z-z_0}{1-\bar{z_0}z}\right)^m.$ Then $f$ satisfies $|f(z)|<1$ for $|z|<1$ and $ f(z)$ has a zero of order $m$ at $z_0$. But $|f(0)|=|z_0|^m$, not $|f(0)|<|z_0|^m$. @Danny – ts375_zk26 Apr 05 '17 at 05:02
  • I feel totally confused now. I`ll leave it for a while then come back to it later. Thanks. – Danny Apr 05 '17 at 05:36