# Showing the holomorphic function contains at least one zero

Let $$D\in \mathbb{C}$$ be an open set and $$f:D\to \mathbb{C}$$ be holomorphic. Suppose we fix $$z_0\in D$$ then we know that there is a local Taylor's expansion in $$B(z_0,r)$$ for some $$r>0.$$ (the corresponding closed ball also lies in $$D$$.)

The question asks if $$|f(z_0)|<\min_{z\in\partial B(z_0,r)}|f(z)|$$ then it must contain a zero in the aforementioned open ball.

I thought to prove by contradiction by saying that $$\frac{1}{f}$$ must be holomorphic on this open ball then I could also deduce the inequality $$|\frac{1}{f(z_0)}|<\max_{z\in\partial B(z_0,r)}|\frac{1}{f(z)}|.$$

However I am not too sure how to proceed further. This kind of reminds me of Rouches Theorem but I am not sure how to apply it.

• OP is using MMP to $\frac 1 f$ so there is no mistake in the stated inequality. Aug 12, 2020 at 23:44
• My answer is correct, there is a mistake in the inequality since he is starting from $f(z_0) < \min f(z)$. Aug 13, 2020 at 13:10
Maximum Modulus Principle shows that if there is no zero then $$\frac 1 {|f(z_0)|} \leq \max_{|z-z_0| \leq r} \frac 1 {|f(z)|}=\max_{|z-z_0| = r} \frac 1 {|f(z)|}$$ But this is equivalent $$|f(z_0| \geq \min \{|f(z)|: |z|= r\}$$. This is a contradiction and hence $$f$$ must have a zero.