$f$ holomorphic in $\mathbb{D}$. Prove $f$ has a zero in $\mathbb{D}$ 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. 

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 $f$.
