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Let $f$ be holomorphic in $\mathbb{D}$ and $f$ be continuous on $\overline{\mathbb{D}}$. Assume $f(0)=c$ and $|f(z)|>|c|$ for $|z|=1$. Prove that $f$ has a zero in $\mathbb{D}$.

Since it's dealing with the number of zeros (or existence of), My initial thought is to find another function $g(z)$ such that $|g(z)-f(z)|<|g(z)|$ for all $z\in\partial\mathbb{D}$, then show that $g$ has at least one zero in $\mathbb{D}$ and use Rouche's theorem to complete the proof. Another approach I thought of is using Argument principle, and showing $\displaystyle\int\dfrac{f'}{f}\geq1$. But I maybe in completely wrong track.

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Just apply Rouche's theorem: $f(z)- c = f(z) + (-c)$ and $|f(z)| > |-c|$ on $|z| = 1.$ Therefore $f(z)-c, f(z)$ have the same number of zeros in $|z|<1$. As $f(0) =c, $ $f(z)$ has at least one zero in the disc.

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Apply the minimum principle: $\lvert f\rvert$ must have a minimum somewhere, but it can't be attained at a $\omega$ such that $\lvert\omega\rvert=1$ (because $\bigl\lvert f(w)\bigr\rvert>\bigl\lvert f(0)\bigr\rvert$. Therefore, it is attained at some $\omega$ with $\lvert\omega\rvert<1$. Therefore, by the minimum principle, $\omega$ is a zero of $f$.

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  • $\begingroup$ Doesn't this only show that $f(z)$ is constant? $\endgroup$ – Ya G Dec 21 '18 at 21:10
  • $\begingroup$ Not at all. It proves that $f$ is constant or it has a zero. But $f$ cannot be constant. So… $\endgroup$ – José Carlos Santos Dec 21 '18 at 21:47
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Hint: Suppose $f$ has no zero in $\mathbb D.$ Consider $1/f.$

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