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Suppose that $f$ is holomorphic inside and on $\gamma(0;1)$, the circle centered at $0$ of radius $1$, with Taylor expansion $\sum_{n=0}^{\infty} c_nz^n$. Given that $f$ has $m$ zeros inside $\gamma(0;1)$, prove that $$\min \{ |f(z)| : |z|=1 \} \leq |c_0| + |c_1|+ \dots + |c_m|$$

I really don't know how to start. From the hypothesis we know that

$$\frac{1}{2\pi i}\int_{\gamma(0;1)}\frac{f'(z)}{f(z)} dz =m$$ but I'm not sure if that fact is useful at all. Maybe I should exploit the series expansion of $f(z)$, but I haven't come up with anything. Any help will be highly appreciated, and thanks in advance!

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  • $\begingroup$ What does $\gamma(0,1)$ stand for? Is it the open disk of unit radius centred at the origin? $\endgroup$ – Arian Jul 29 '16 at 17:21
  • $\begingroup$ @Arian It is the circle centerd at zero with radius $1$, I'll edit the question to make that clearer. $\endgroup$ – user313212 Jul 29 '16 at 17:27
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I thank very much @See-Woo Lee who has indicated the right way to prove the issue, i.e., by contradiction.

Let $f(z)=p(z)+q(z)$ with $\begin{cases}p(z)&=&\sum_{n=0}^{m}c_nz^n\\q(z)&=&\sum_{n=m+1}^{\infty} c_nz^n \end{cases}$.

Let $U$ be the unit disk and $\partial U$ be the unit circle.

Let us prove the result by contradiction.

Let us assume the contrary of what has to be proven, i.e., that

$$\forall z \in \partial U, \ |f(z)| > |c_0| + |c_1|+ \dots + |c_m|$$

thus, a fortiori

$$\forall z \in \partial U, \ |f(z)| > |c_0 + c_1 z+ \dots + c_m z^m|=|p(z)|=|f(z)-q(z)|$$

Using Rouché's theorem, $f(z)$ and $f(z)-q(z)=p(z)$ have the same number of zeros inside the unit disk $U$, i.e., $m$ zeros.

But $q(z)=z^{m+1}r(z)$ where $r(z)$ is a holomorphic function; therefore $q(z)$ would have at least $m+1$ zeros in $U$, a contradiction with the previous sentence.

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    $\begingroup$ can you expand the reasoning ? $\endgroup$ – Hamza Jul 29 '16 at 17:36
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    $\begingroup$ So I should proceed by comtradiction? Assume that $\vert \sum_{n=m+1}^{\infty} c_nz^n \vert < |f(z)|$. Then $g(z)$ has $m$ zeros inside the contour $\gamma(0;1)$, but I can't see any contradiction there. $\endgroup$ – user313212 Jul 29 '16 at 19:52
  • $\begingroup$ I have provided a thorough proof now ; yes, you were on the good track by working by contradiction. $\endgroup$ – Jean Marie Jul 29 '16 at 22:01
  • $\begingroup$ Thanks, @Hamza, for the correction of "polynomial" to "holomorphic". $\endgroup$ – Jean Marie Jul 30 '16 at 0:10
  • $\begingroup$ excuse me because i do it without telling you $\endgroup$ – Hamza Jul 30 '16 at 0:18

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