If $|f(z)|>0$ then $f$ has no zeros.

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I am trying to understand the proof of Rouche's Theorem.

Rouche's Theorem

Let $C$ denote a simple closed contour, and suppose that

a) two functions f(z) and g(z) are analytic on and inside $C$

b) $|f(z)|>|g(z)|$ at each point on $C$

Then $f(z)$ and $f(z)+g(z)$ have the same number of zeroes, counting multiplicities inside $C$.

In the proof it was mentioned that $f(z)$ has no zero in $C$ since $|f(z)|>0$ at each point in $C$.

Is this true in general? I tried to make a statement out of it:

If $f$ is defined in a domain $S\subseteq \mathbb{C}$ and $|f(z)|>0$ for all $z\in S$ then $f$ has no zero in $S$.

Proof: Suppose $f$ has a zero $w$ in $S$. $f(w)=0\implies |f(w)|=0$ which is a contradiction since $|f(z)|>0 \,\forall z\in S$.

Is something wrong with my statement above? Does it need $f$ to be analytic? Or is it wrong in general? Thank you!

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This is a statement about individual values of a function, i.e., about complex numbers. It is true that a complex number is zero if and only if its modulus is zero. More formally: for $w\in\mathbb C$, we have $w\ne 0$ if and only if $|w|>0$. (As D.F. wrote).