# $f$ 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.

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