A confuse about big Rudin's remark I am confused about a remark made in Rudin's book RCA 10.10
The remark says that

$2\pi$$Ind_\gamma(z)$ is the net increase in the imaginary part of $\lambda(t)$,as t runs from $\alpha$ to $\beta$,and this is the same as the net increase of the argument of $\gamma(t)-z$.If we divide this increase by $2\pi$,we obtain “the number of times that $\gamma$ winds around $z$”.

Where,$\gamma$ is a closed path,$$\lambda(t)=\int_\alpha^t \frac{\gamma’{s}} {\gamma(s)-z}ds$$,and $$Ind_\gamma(z)=\frac{1}{2\pi i}\int_\alpha^\beta \frac{\gamma’(s)}{\gamma(s)-z}ds$$.
My confusion is that I don’t know the meaning of “net increase”.  And how can we obtain “the number of times that...”
 A: This belongs to the basics of complex analysis.
Consider two points $z_j\ne0$ $\>(j=0,1)$ near each other, e.g. in the same half plane $H$, and let $\gamma\subset H$ be a  path connecting $z_0$ with $z_1$. This (short) $\gamma$ does not wind around the origin. Then
$$\int_0^1{\gamma'(t)\over\gamma(t)}\>dt=\int_\gamma{dz\over z}=\log{|z_1|\over|z_0|}+i\>\Delta\phi\ ,$$
where $\phi$ is a polar angle valid around the place, and
$$\Delta\phi=\arg(z_1)-\arg(z_0)\ \in\>]{-\pi},\pi[$$
is the small increase of the polar angle along $\gamma$. This increase may of course be negative; at any rate it has only to do with the position of the $z_j\in H$.
If you now have a long closed $\gamma$ that  might wind a few times around the origin you can choose anchor points $z_j\in\gamma$ $\>(0\leq j\leq N, \>z_N=z_0)$. In this way $\gamma$ is partitioned into $N$ arcs $\gamma_j$ of the kind considered above, connecting $z_{j-1}$ with $z_j$. We now have
$$\int_\gamma{dz\over z}=\sum_j\int_{\gamma_j}{dz\over z}=\sum_j\log{|z_j|\over|z_{j-1}|}+i \sum_j(\Delta\phi)_{\gamma_j}=0+i \sum_j(\Delta\phi)_{\gamma_j}\ .$$
It turns out that $\int_\gamma{dz\over z}$ is $i$ times the sum of all small increases (decreases) of the polar angle along $\gamma$. This sum is $2\pi$ times the number ${\rm Ind}(\gamma,0)$ of positive windings of $\gamma$ around $0$, because each such winding increases the polar angle totally by $2\pi$.
