Correlation between winding number and degree. Let $\mathbb{C}^*=\mathbb{C} \setminus\{0\}$. Suppose $f:S^1\rightarrow\mathbb{C}^*$ is a continuous map such that the composite function $\alpha=f\circ\pi:[0,1]\rightarrow\mathbb{C}^*$ is piecewise continuously differentiable. The winding number of $f$ is defined as the complex number
\begin{align*}
w(f)=\frac{1}{2\pi i}\int_\gamma\frac{dz}{z}=\frac{1}{2\pi i}\int_0^1\frac{\alpha'(s)}{\alpha(s)}ds
\end{align*}
where the curve $\gamma=f(S^1)$ is oriented by $\alpha.$ If $g(z)=f(z)/|f(z)|$ for all $z\in S^1$, prove that $w(f)=deg(g)$.
The only additional information I am given is that I am allowed to assume a theorem from complex analysis saying that the value of $\int_\gamma\frac{dz}{z}$ depends only on the homotopy class of $f$.
I am at a loss on this exercise. I've just barely covered the degree of a map from $S^1$ to $S^1$ and it seems as though this problems has so many components that I'm having trouble seeing how to break it down. Any help is greatly appreciated.
 A: First of all, we have the following composition of functions
$t\rightarrow π(t)\rightarrow f(π(t))\rightarrow \frac{f(π(t))}{|f(π(t))|}$
$[0,1]\rightarrow S^1 \rightarrow C^{*}\rightarrow S^1$
where $π(t)=e^{2πit}$, $f(π(t))=α(t)$, $g(π(t))=\frac{f(π(t))}{|f(π(t))|}$, $t\in [0,1]$ .
In the exercise above we are talking about the winding number of $f$, but it is preferable to talk about the winding number of the curve $α$ defined as above:
$w(α)=\frac{1}{2πi}\int_{0}^{1}\frac{α'(s)}{α(s)}ds$
CLAIM 1: $\int_{0}^{1}\frac{α'(s)}{α(s)}ds=2kπi, k\in Z$, so $w(α)=k$.
[PROOF: Let $φ:[0,1]\rightarrow C$, with $φ(t)=exp(\int_{0}^{t}\frac{α'(s)}{α(s)}ds)$.
Then $φ'(t)=φ(t)\frac{α'(s)}{α(s)}$ which gives $(\frac{φ(t)}{α(t)})'=0$, so the function $\frac{φ(t)}{α(t)}$ is constant. Also, $φ(0)=1$ and
$\frac{φ(1)}{α(1)}=\frac{φ(0)}{α(0)}=\frac{1}{α(0)}$
which gives, since $α(0)=α(1)$, (α is a closed curve),
$φ(1)=1$.
That means $exp(\int_{0}^{1}\frac{α'(s)}{α(s)}ds)=1$, so $\int_{0}^{1}\frac{α'(s)}{α(s)}ds=2kpi$ for some integer $k$.]
We set $γ(t)=g(π(t))=\frac{α(t)}{|α(t)|}$ which is a curve from $[0,1]$ to $S^1$. We assume without harm $γ(0)=1$. So, we may write $γ(t)=e^{2πiΦ(t)}=(cos2πΦ(t),sin2πΦ(t))=(γ_{1}(t),γ_{2}(t))$. Here $Φ(t)$ is the lift $Φ: [0,1]\rightarrow R$ of $γ$ to $R$. At the same time $g(e^{2πit})=γ(t)=e^{2πiΦ(t)}$, so $Φ$ may be considered also as the lift of $g$ to $R$.
CLAIM 2: $Φ(1)=k=w(α)$
[PROOF: Let $α(t)=α_{1}(t)+iα_{2}(t)$. Then,
$\frac{1}{2πi}\frac{α'}{α}=$$\frac{1}{2πi}\frac{(α_{1}'+iα_{2}')(α_{1}-iα_{2})}{|α|^{2}}=$
$\frac{1}{2π|α|^{2}}(-α_{2}α_{1}'+α_{1}α_{2}'-i(α_{1}α_{1}'+α_{2}α_{2}'))$. Thus,
$\frac{1}{2πi}\int_{0}^{1}\frac{α'}{α}=\frac{1}{2π}\int_{0}^{1}\frac{-α_{2}α_{1}'+α_{1}α_{2}'}{|α|^{2}}-\frac{i}{2π}\int_{0}^{1}\frac{α_{1}α_{1}'+α_{2}α_{2}'}{|α|^{2}}$.
The second integral of the right member of the above equation is zero, since the curve $α$ is closed.It is also easy to compute the first integral of the right member and get actually
$\frac{1}{2πi}\int_{0}^{1}\frac{α'}{α}=\frac{1}{2π}\int_{0}^{1}\frac{-α_{2}α_{1}'+α_{1}α_{2}'}{|α|^{2}}=\frac{1}{2π}\int_{0}^{1}(-γ_{2}γ_{1}'+γ_{1}γ_{2}')=$
$\frac{1}{2π}\int_{0}^{1}[-sin2πΦ(t)(-2π)Φ'(t)sin2πΦ(t)+cos2πΦ(t)2πΦ'(t)cos2πΦ(t)]dt=$
$\int_{0}^{1}Φ'(t)dt=Φ(1)-Φ(0)=Φ(1)$.]
CLAIM 3: $deg(g)=Φ(1)$
[PROOF:From above we have that $Φ:[0,1]\rightarrow R$ is also the lift of $g:S^1\rightarrow S^1$, namely, $g(e^{2πit})=e^{2πiΦ(t)}$.
To find $deg(g)$ we need to find first the number of points in the preimage $g^{-1}(1)$. Actually, we seek all $t\in [0,1]$ such that $g(e^{2πit})=1=g(1)=g(e^{2π0})$. Equivalently,
$Φ(t)-Φ(0)$ is integer number.      (1)
The numbers $0$ and $1$ are such, and we will show there are $k-2$ more such numbers. But this is obvious by (1) and the fact that  $Φ(1)-Φ(0)=k$ shows there exist only k "circles" between $Φ(1)$ and $Φ(0)$.
Now on every of its values, thus on $1\in S^1 $ too,  $g$ is regular. Thus, on each of the previous $k$ preimages $g$ is a local diffeomorphism. On each of them preserves the orientation or on each of them does not, with index $+1$ or $-1$ respectively. Thus, $deg(g)=k$ or $-k$. Assuming $g$ preserves orientation, the proof is finished.]
REMARK:If we consider the form $η=\frac{1}{2π}[\frac{-ydx}{x^{2}+y^{2}}+\frac{xdy}{x^{2}+y^{2}}]$, which corresponds, through the metric, to the gradient field of the function $arctan(\frac{y}{x})$, then we have already proven that
$\int_{γ([0,1])}η=\int_{0}^{1}η(γ'(t))dt=Φ(1)=k$.
This is the way we measure the number of rotations around $0$ of a closed plane curve $γ$.
