Topological degree of a complex valued map defined over a circle 
Given a continuous map $f \colon S^n \to S^n$, it induces a map $f_{*} \colon \tilde{H}_n(S^n) \to \tilde{H}_n(S^n)$ of the form $f_{*}(z)=k*z$, where $k$ is an integer. Define the degree of $f$ as $k$.

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Let $$p(z) = a_nz^n+a_{n-1}z^{n-1} + \dots +a_0 $$ a complex valued polynomial function. Let $\gamma : C \to S^1$ $$\gamma(z) = \dfrac{p(z)}{\mid p(z)\mid}$$  where $C$ is a circle contained in $\mathbb{C}$ (the same for $S^1$). Obviously $C$ doesn't touch any roots, so $\gamma$ is "well defined".
1) I have to prove that if the circle contains only one root of $p$ (call it $\alpha$), then the topological degree is $m$, the multiplicity of such root of $p$. 

My Attempt. let $p(z)=(z-\alpha)^m\tilde{p}(z)$. Then consider the homotopy $$\Gamma(s,z) := \dfrac{ \tilde{p}(s\alpha + (1-s)z)(z-\alpha)^m}{\mid \tilde{p}(s\alpha + (1-s)z)(z-\alpha)^m \mid}$$ where $s$ ranges in $[0,1]$ and then I have to prove that $\dfrac{(z-\alpha)^m}{\mid (z-\alpha)^m \mid}$ has degree $m$.

2) Second task is to prove that if the circle contains two roots $\alpha, \beta$ then the degree of the map $\gamma$ is the sum of the multiplicities ($m,n$) of such roots.
Here is my attempt I'm unsure about the correctness of such homotopy. Can someone help me pointing out why it is correct (or incorrect?)

consider the homotopy $$ H(z,t) := := \dfrac{ \tilde{p}(s\alpha + (1-s)z)(z-\alpha)^m(z-[(1-s)\beta+sa])^n}{\mid  \tilde{p}(s\alpha + (1-s)z)(z-\alpha)^m(z-[(1-s)\beta+sa])^n \mid}$$

The doubts arise because it's not very clear to me when an homotopy is valid and when not. Here I think it is right, because it is obviously continuous, but the fact that my polynomial is defined over a circle (so there is a hole in between) doesn't let me to be sure of such function. 
 A: First of all, the fact that $(z-\alpha)^n$ has degree $n$ over $C$ is equivalent to prove that $z^n$ has degree $n$ over a small circle around the origin. Intuitively $z^n$ wraps around the circle $n$ times. This can be formalized using the concepts of local degree and the formula for calculating the degree of a map from its local degrees (see for example Hatcher vol.1 page 136-137)
Then for the second task, we can consider the homotopy $$ H(z,t) := := \dfrac{ \tilde{p}(s\alpha + (1-s)z)(z-\alpha)^m(z-[(1-s)b+sa])^n}{\mid  \tilde{p}(s\alpha + (1-s)z)(z-\alpha)^m(z-[(1-s)b+sa])^n \mid}$$
Which is well defined thanks to the assumption on $p$ and $\tilde{p}$. (in particular the denominator does not vanish (the other roots are outside the circle and so $\tilde{p}(s\alpha + (1-s)z) \neq 0$ for every $s \in [0,1]$. To conclude, just note that the degree of $(z-\alpha)^{m+n}$ is obviously $m+n$
It is important to notice that with this reasoning I can't move roots from the outside in the interior of the circle, (the homotopy in the intersection point between the segment and the circle is not defined)
I have to thanks Daniel Fischer for the "review" of my attempt
