Beardon's book is sadly out of print, but detailed proofs of results
concerning the winding number, or the index [incidentally, I
don't think that one can distinguish these two terms, at least not
in the way attempted in the question, although the same number
associated with a plane curve is defined in the literature in two
different ways, and the question is therefore an interesting one],
can also be found in Chapter 3 of Ash and Novinger,
Complex Variables (2004), which is downloadable
here.
Because of the importance of the concept, I think it is worth giving
a fairly short and straightforward proof from first principles,
rather than appealing to a series of general theorems (although that
is admittedly more efficient, and it would probably also be worth
giving a direct proof of the "fundamental theorem of calculus" for
rectifiable paths).
Digression:
This is especially so as it is not easy even to find books that
bother to treat the case of rectifiable curves. The best-known
exception may be Conway, Functions of One Complex Variable I
(2nd ed. 1978). Another exception is
Garling, A Course in Mathematical Analysis, Vol. III (2014).
But even these two books prove
the equality in question by using piecewise continuously
differentiable approximations to rectifiable curves; and so it is
hard, for me at least, to see the wood for the trees.
Rudin, in Principles of Mathematical Analysis (3rd ed. 1976),
remarks of the concept of rectifiability:
"The case $k = 2$ (i.e., the case of plane curves) is of
considerable importance in the study of analytic functions of a
complex variable."
In contrast, in Real and Complex Analysis (3rd ed. 1987),
the same author writes:
"In this section the required integration theory will be developed;
we shall keep it as simple as possible, and shall regard it merely
as a useful tool in the investigation of properties of holomorphic
functions. [...] A path is a piecewise continuously
differentiable curve in the plane."
It the latter, more utilitarian view of curves that seems to
dominate the literature on complex analysis. (There are some remarks
about this in Burckel, An Introduction to Classical Complex
Analysis, Vol. 1; but I haven't got that book yet, so I can't say
any more about it.)
(End of amateur lecture.) :)
Let $[a, b]$ be a compact interval of $\mathbb{R}$, and
$\gamma \colon [a, b] \to \mathbb{C}$ a continuous function. The
image set $[\gamma] = \gamma([a, b])$ is a compact subset of
$\mathbb{C}$, therefore it is closed.
Suppose $0 \notin [\gamma]$. Because $[\gamma]$ is closed, there
exists $R > 0$ such that the open disc $\mathbb{C}(0, R)$, with
centre $0$ and radius $R$, does not intersect $[\gamma]$.
Because $\gamma$ is a continuous function on a compact set, it is
uniformly continuous. So there exists a partition of $[a, b]$,
\begin{equation}
\tag{1}\label{eq:3257640:part:2}
a = t_0 < t_1 < \cdots < t_{n-1} < t_n = b,
\end{equation}
such that
$$
|\gamma(t) - \gamma(t')| < R \quad
(t_{j-1} \leqslant t < t' \leqslant t_j, \
1 \leqslant j \leqslant n).
$$
Therefore, each of the sets $\gamma([t_{j-1}, t_j])$ is contained in
an open disc (of radius $R$) not containing the origin. Because
there exists a continuous choice of logarithm on any such disc, a
simple inductive proof shows that
there exists a continuous function
$\lambda \colon [a, b] \to \mathbb{C}$ such that
$\gamma(t) = e^{\lambda(t)}$ for all $t \in [a, b]$.
Recall that $e^z = 1 + z + \rho(z)$, where $\rho(z)/z \to 0$ as
$z \to 0$.
Take $\epsilon > 0$ such that $|\rho(z)| < |z|/2$ when
$0 < |z| < \epsilon$. (For example, we could take
$\epsilon = \tfrac{3}{4}$. Not that it matters, of course!)
Because $\lambda$ is uniformly continuous, there exists
$\delta > 0$ such that
$|\lambda(t) - \lambda(t')| < \epsilon$ when $|t - t'| < \delta$.
Therefore,
\begin{align*}
|\gamma(t) - \gamma(t')| & = |e^{\lambda(t)} - e^{\lambda(t')}|
\\ & = |e^{\lambda(t')}||e^{\lambda(t) - \lambda(t')} - 1|
\\ & = |\gamma(t')||\lambda(t) - \lambda(t') +
\rho(\lambda(t) - \lambda(t'))|
\\ & \geqslant \frac{R|\lambda(t) - \lambda(t')|}{2}
\quad (|t - t'| < \delta).
\end{align*}
In a partition \eqref{eq:3257640:part:2} (not necessarily with the
same values of $t_j$ as before, of course),
if $|t_j - t_{j-1}| < \delta$ ($j = 1, \ldots, n$), then
$$
\sum_{j=1}^n|\lambda(t_j) - \lambda(t_{j-1})| \leqslant
\frac{2}{R}\sum_{j=1}^n|\gamma(t_j) - \gamma(t_{j-1})|.
$$
If $\gamma$ is rectifiable, and has length $L \geqslant 0$, it
follows that $\lambda$ is rectifiable, and has length at most
$2L/R$.
The path integral
\begin{equation}
\tag{2}\label{eq:3257640:path:2}
\int_\gamma \frac{dz}{z} = \int_a^b \frac{d\gamma(t)}{\gamma(t)}
\end{equation}
exists whenever the complex Riemann-Stieltjes integral on the right
is defined. (For a general definition of complex Riemann-Stieltjes
integrals, see e.g. Apostol, Mathematical Analysis (2nd ed. 1974).)
Given a partition \eqref{eq:3257640:part:2} - again, it has no
necessary relation to previous such partitions - "tagged", as they
say, with values $s_j \in [t_{j-1}, t_j]$ ($j = 1, \ldots, n$), the
Riemann-Stieltjes sum for the integral on the right of
\eqref{eq:3257640:path:2} is:
\begin{align}
& \phantom{={}}
\sum_{j=1}^n\frac{\gamma(t_j) - \gamma(t_{j-1})}{\gamma(s_j)}
\notag \\ & =
\sum_{j=1}^n\frac{e^{\lambda(t_j)} - e^{\lambda(t_{j-1})}}
{e^{\lambda(s_j)}}
\notag \\ & =
\sum_{j=1}^n\left(e^{\lambda(t_j) - \lambda(s_j)} -
e^{\lambda(t_{j-1}) - \lambda(s_j)}\right)
\notag \\ & =
\sum_{j=1}^n\left[\lambda(t_j) - \lambda(t_{j-1}) +
\rho(\lambda(t_j) - \lambda(s_j)) -
\rho(\lambda(t_{j-1}) - \lambda(s_j))\right]
\notag \\ & =
\tag{3}\label{eq:3257640:sum:log}
[\lambda(b) - \lambda(a)] +
\sum_{j=1}^n\rho(\lambda(t_j) - \lambda(s_j)) -
\sum_{j=1}^n\rho(\lambda(t_{j-1}) - \lambda(s_j)).
\end{align}
To prove that the Riemann-Stieltjes integral
\eqref{eq:3257640:path:2} exists, and that it is equal to
$\lambda(b) - \lambda(a)$, it is enough (see e.g. Exercise 7.3 of
Apostol) to prove that for every $\epsilon' > 0$ there exists
$\delta' > 0$ such that if $t_j - t_{j-1} < \delta'$
($j = 1, \ldots, n$), then
\begin{equation}
\tag{4}\label{eq:3257640:integral}
\left\lvert
\sum_{j=1}^n\frac{\gamma(t_j) - \gamma(t_{j-1})}{\gamma(s_j)} -
[\lambda(b) - \lambda(a)]
\right\rvert
< \epsilon'.
\end{equation}
If $L = 0$, i.e. if $\gamma$ is constant (a point "curve"), then
$\lambda$ is also constant, and \eqref{eq:3257640:integral} is
trivially satisfied; so, suppose from now on that $L > 0$; then at
least some of the "error" terms in \eqref{eq:3257640:sum:log} are
non-zero. Let $\eta > 0$ be such that
$$
\frac{|\rho(z)|}{|z|} < \frac{R\epsilon'}{2L}
\quad (0 < |z| < \eta).
$$
Let $\delta' > 0$ be such that
$$
|\lambda(t) - \lambda(t')| < \eta \quad (|t - t'| < \delta').
$$
Then, for any partition \eqref{eq:3257640:part:2} satisfying
$t_j - t_{j-1} < \delta'$ ($j = 1, \ldots, n$), tagged with values
$s_j \in [t_{j-1}, t_j]$ ($j = 1, \ldots, n$), the "error" in
\eqref{eq:3257640:sum:log} is bounded above, thus:
\begin{align*}
& \phantom{={}}
\left\lvert\sum_{j=1}^n\rho(\lambda(t_j) - \lambda(s_j)) -
\sum_{j=1}^n\rho(\lambda(t_{j-1}) - \lambda(s_j))\right\rvert
\\ & \leqslant
\sum_{j=1}^n\left(|\rho(\lambda(t_j) - \lambda(s_j))| +
|\rho(\lambda(t_{j-1}) - \lambda(s_j))|\right)
\\ & <
\frac{R\epsilon'}{2L}
\sum_{j=1}^n\left(|\lambda(t_j) - \lambda(s_j)| +
|\lambda(t_{j-1}) - \lambda(s_j)|\right)
\\ & \leqslant \epsilon'.
\end{align*}
We have thus proved
$$
\int_\gamma \frac{dz}{z} = \lambda(b) - \lambda(a),
$$
on the assumption only that $\gamma$ is rectifiable. If also
$\gamma$ is closed, i.e. $\gamma(a) = \gamma(b)$, then
$\operatorname{re}\lambda(a) = \operatorname{re}\lambda(b)$, and
$\operatorname{im}\lambda(t)$ is a choice of
$\operatorname{arg}\gamma(t)$ for $t \in [a, b]$, therefore:
$$
\int_\gamma \frac{dz}{z} = 2\pi{i}n(\gamma, 0),
$$
where I have used what seems to be the nearest thing to a standard
notation for the index, or winding number, of $\gamma$ about $0$.
It’s a consequence of the following lemma: let $\gamma:[0,1] \rightarrow \mathbb{C}^*$ be a closed curve. There exists $\epsilon >0$ such that for every closed curve $\delta:[0,1] \rightarrow \mathbb{C}^*$ with $\|\gamma-\delta\|_{\infty} <\epsilon$, $I(\gamma,0)=I(\delta,0)$.
Indeed, you get your result by taking $\gamma$ and $\delta$ to be close translates of the same closed curve.
How to prove the lemma? Well, let $r>0$ be such that $|\gamma| \geq r$, and assume $\|\gamma-\delta\|_{\infty} < r$. Then consider $\alpha=\gamma/\delta$: then $\alpha([0,1]) \subset D=\mathbb{C} \backslash (-\infty,0]$.
But it’s easy to see that there is a complex logarithm $L: D \rightarrow \mathbb{C}$, and then the imaginary part of $L \circ \alpha$ is an argument of $\alpha$. Therefore $\mathrm{vararg}(\alpha)=0$, and thus $\mathrm{vararg}(\gamma)=\mathrm{vararg}(\delta)$ so $I(\gamma,0)=I(\delta,0)$.
Best Answer
The condition $\operatorname{Re}\frac{\gamma(t)}{\gamma_{0}(t)} > 0$ means $\gamma(t)$ lies in the same (light gray) open half-plane from the origin as $\gamma_{0}(t)$.
The condition $$ |\gamma(t) - \gamma_{0}(t)| < d = \min \{|\gamma_{0}(t)| : 0 \leq t \leq 1\} $$ means $\gamma(t)$ lies in the (darker gray) disk, which is contained in the open half-plane above.
If $|a(0) - a_{0}(0)| < \pi/2$, i.e., the arguments of $\gamma(0)$ and $\gamma_{0}(0)$ agree within $\pi/2$, then since $\gamma_{0}(t)$ and $\gamma(t)$ lie in the same half-plane through $0$ for all $t$, we have $|a(t) - a_{0}(t)| < \pi/2$ for all $t$.