Let $\gamma:[0,1] \rightarrow \mathbb{C}$ be a continuous map never taking the value 0 and satisfying $\gamma(0)=\gamma(1)$. Define the degree (or winding number) $w(\gamma ; 0)$ of $\gamma$ about 0 . Prove the following.

(i) If $\delta:[0,1] \rightarrow \mathbb{C} \backslash\{0\}$ is a continuous map satisfying $\delta(0)=\delta(1)$, then the winding number of the product $\gamma \delta$ is given by $w(\gamma \delta ; 0)=w(\gamma ; 0)+w(\delta ; 0)$.

(ii) If $\sigma:[0,1] \rightarrow \mathbb{C}$ is continuous, $\sigma(0)=\sigma(1)$ and $|\sigma(t)|<|\gamma(t)|$ for each $0 \leqslant t \leqslant 1$, then $w(\gamma+\sigma ; 0)=w(\gamma ; 0)$.

(iii) Let $D=\{z \in \mathbb{C}:|z| \leqslant 1\}$ and let $f: D \rightarrow \mathbb{C}$ be a continuous function with $f(z) \neq 0$ whenever $|z|=1$. Define $\alpha:[0,1] \rightarrow \mathbb{C}$ by $\alpha(t)=f\left(e^{2 \pi i t}\right)$. Then if $w(\alpha ; 0) \neq 0$, there must exist some $z \in D$, such that $f(z)=0$. [It may help to define $F(s, t):=f\left(s e^{2 \pi i t}\right)$. Homotopy invariance of the winding number may be assumed.]