Theorem 1:
Suppose that $\Gamma \subseteq \mathbb{C}\backslash\{0\}$ is a closed contour. If $|c|<d(\Gamma,0)$, then $W(\Gamma,c)=W(\Gamma,0)$.
This theorem can be used to show that the number of zeroes (counting multiplicity) of a homomorphic function will locally stay the same:
Theorem 2:
Suppose that $f$ is holomorphic in $z_0$ with $0<\underset{z=z_0}{\operatorname{mult}}f(z)=:m<\infty$. Then, for each small $\epsilon >0$, there exists a $\delta>0$ such that $f$ reaches the value $c$ exactly $m$ times in $B(z_0,\epsilon)$, for each $c\in B'(0,\delta)$.
($B'(z,r)$ is a punctured neighbourhood of $z$, i.e. $B(z,r)\backslash \{z\}$).
Proof:
Since $m<\infty$, we can conclude that $z_0$ is an isolated zero (otherwise $f=0$ in a neighbourhood of $z_0$). Therefore, we can find an $\epsilon_0>0$ such that $f$ is a holomorphic function on $B(z_0,\epsilon_0)$ and has no zeroes in $B'(z_0,\epsilon_0)$. Let $\epsilon < \epsilon_0$ and $\Gamma:=\partial B(z_0,\epsilon)$. It is clear that $\Gamma$ has no zeroes of $f$. Since $f(\Gamma)$ is compact and $f(\Gamma)\cap\{0\}=\emptyset$, we see that $\delta:= d(f(\Gamma),0)>0$. Theorem 1 implies that $W(f(\Gamma_+),c)=W(f(\Gamma_+),0)$ for each $c\in \mathbb{C}$ with $|c|<\delta$. $\color{red}{\text{The argument principle implies that the number of zeroes of}}$ $\color{red}{f(z)-c \,\,\text{in}\,\, B(z_0,\epsilon) \,\,\text{is equal to the number of zeroes of}\,\, f \,\,\text{in}\,\, B(z_0,\epsilon) \,\,\text{(counting multiplicity), i.e. equal to }\,\, m.}$
If $m=1$, then $\color{red}{\text{the theorem follows immediately}}$. Suppose now that $m>1$, then $f'(z_0)=0$, we choose $\epsilon_0$ such that $f'$ has no zeroes in $B'(z_0,\epsilon_0)$. Then $\color{red}{\text{each zero of}\,\, f(z)-c \,\,\text{in}\,\, B'(z_0,\epsilon_0)\,\,\text{is simple}.}$ This implies that $f(z)-c$ has $m$ different zeroes in $B(z_0,\epsilon_0)$ when $c\ne 0$.
I'm having trouble understanding the red parts. Could someone explain these to me.
Thank you.