If \[f(z)=\sum_{n=0}^{\infty} c_{n}(z-a)^{n} \quad \text{ for } \quad z \in D(a ; R) \tag{*}\] and if \(0<r<R\), then
\[\sum_{n=0}^{\infty}\left|c_{n}\right|^{2} r^{2 n}=\frac{1}{2 \pi} \int_{-\pi}^{\pi}\big|f\big(a+r e^{i \theta}\big)\big|^{2} d \theta \tag{**}\]
Proof: We have
\[f\big(a+r e^{i \theta}\big)=\sum_{n=0}^{\infty} c_{n} r^{n} e^{i n \theta}.\] For \(r<R\), the series converges uniformly on \([-\pi, \pi]\).
Since \[\frac{1}{2 \pi} \int_{-\pi}^{\pi} e^{-i n \theta} d \theta= \begin{cases} 1 & \text{ if } n=0,\\ 0 & \text{ if } n\in\mathbb Z\setminus\{0\}, \end{cases}\] we conclude that \[c_{n} r^{n}=\frac{1}{2 \pi} \int_{-\pi}^{\pi} f\big(a+r e^{i \theta}\big) e^{-i n \theta} d \theta,\] and \[\begin{aligned} \frac{1}{2 \pi} \int_{-\pi}^{\pi}\big|f\big(a+r e^{i \theta}\big)\big|^{2} d \theta &=\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}c_n\overline{c_m}r^{n+m}\frac{1}{2 \pi} \int_{-\pi}^{\pi} e^{i (n-m) \theta} d \theta\\ &=\sum_{n=0}^{\infty}\left|c_{n}\right|^{2} r^{2 n}. \end{aligned}\]
This completes the proof. $$\tag*{$\blacksquare$}$$
Let \(f\) be defined on \(\Omega\) and \(a \in \Omega\). Then \(|f|\) has a local maximum at \(a\) if there exists \(\delta>0\) such that \(D(a, \delta) \subseteq \Omega\) and \(|f(a)| \geq|f(z)|\) for every \(z \in D(a, \delta)\). Further, we say that \(|f|\) has no local maximum in \(\Omega\) if \(|f|\) does not have local maximum at every point of \(\Omega\). Similarly, we define a local minimum.
Suppose that \(\Omega\) is a region and \(f \in H(\Omega)\).
Then \(|f|\) has no local maximum at any point of \(\Omega\), unless \(f\) is constant.
Moreover, if the closure of \(\Omega\) is compact and \(f\) is continuous on \(\overline{\Omega}\), then \[\sup_{z\in\Omega}|f(z)|\le \sup_{z\in\partial\Omega}|f(z)|.\]
Proof: We now prove part (a). Suppose that \(|f|\) attains a local maximum at \(a\in\Omega\), then \[|f(z)|\le |f(a)| \quad \text{ for all }\quad z\in D(a, \delta)\] for some \(\delta>0\).
If \(z\in D(a, \delta)\), then it can be represented as \[z=a+re^{i\theta}\] for some \(r\in[0, \delta)\) and \(\theta\in[-\pi, \pi]\). Hence \[|f(a+re^{i\theta})|\le |f(a)| \quad \text{ for all }\quad r\in[0, \delta) \text{ and } \theta\in[-\pi, \pi].\]
Since \[f(z)=\sum_{n\ge0}c_n(z-a)^n\quad \text{ for }\quad z\in D(a, \delta).\]
By the previous theorem it follows that \[\sum_{n=0}^{\infty}\left|c_{n}\right|^{2} r^{2 n}=\frac{1}{2 \pi} \int_{-\pi}^{\pi}\big|f\big(a+r e^{i \theta}\big)\big|^{2} d \theta \leq|f(a)|^{2}=\left|c_{0}\right|^{2}.\]
Hence \[\sum_{n=1}^{\infty}\left|c_{n}\right|^{2} r^{2 n}\le 0\] and consequently \(c_{1}=c_{2}=c_{3}=\cdots=0\), which implies that \(f(z)=f(a)\) in \(D(a ; r)\).
Since \(\Omega\) is connected, then \(f\) must be constant in \(\Omega\).
We now prove part (b). There is nothing to prove if \(f\) is constant. Suppose that \(f\) is non-constant in \(\Omega\).
Since \(f\) is continuous on \(\overline{\Omega}\), which is compact, then it attains its maximum in \(\overline{\Omega}\).
Since \(f\) is non-constant this maximum must be attained in \(\partial \Omega\), otherwise \(f\) would be constant by part (a).$$\tag*{$\blacksquare$}$$
Suppose that \(\Omega\) is a region, \(f \in H(\Omega)\), and \(\overline{D}(a ; r) \subseteq \Omega\).
Then
\[|f(a)| \leq \max_{\theta\in[-\pi, \pi]}\big|f\big(a+r e^{i \theta}\big)\big|. \tag{*}\] Equality occurs in (*) if and only if \(f\) is constant in \(\Omega\). Consequently, \(|f|\) has no local maximum at any point of \(\Omega\), unless \(f\) is constant.
Moreover, we have \[|f(a)| \geq \min_{\theta\in[-\pi, \pi]}\big|f\big(a+r e^{i \theta}\big)\big|\] if \(f\) has no zero in \(D(a ; r)\).
If \(f \in H(\Omega)\) and \(g\) is defined in \(\Omega \times \Omega\) by \[g(z, w)= \begin{cases} \frac{f(z)-f(w)}{z-w} & \text { if } w \neq z, \\ f^{\prime}(z) & \text { if } w=z, \end{cases}\] then \(g\) is continuous in \(\Omega \times \Omega\).
Proof: The only points \((z, w) \in \Omega \times \Omega\) at which the continuity of \(g\) is possibly in doubt have \(z=w\).
Fix \(a \in \Omega\), and \(\varepsilon>0\). There exists \(r>0\) such that \(D(a ; r) \subseteq \Omega\) and \[\left|f^{\prime}(\zeta)-f^{\prime}(a)\right|<\varepsilon\]
for all \(\zeta \in D(a ; r)\).
If \(z\) and \(w\) are in \(D(a ; r)\) and if
\[\zeta(t)=(1-t) z+t w,\]
then \(\zeta(t) \in D(a ; r)\) for \(0 \leq t \leq 1\), and
\[g(z, w)-g(a, a)=\int_{0}^{1}\left[f^{\prime}(\zeta(t))-f^{\prime}(a)\right] d t.\]
The absolute value of the integrand is \(<\varepsilon\), for every \(t\in[0, 1]\). Thus \(|g(z, w)-g(a, a)|<\varepsilon\). This proves that \(g\) is continuous at \((a, a)\). $$\tag*{$\blacksquare$}$$
Let \(\Omega\subseteq \mathbb C\) be open. Suppose that \(\varphi \in H(\Omega)\), and \(\varphi^{\prime}\left(z_{0}\right) \neq 0\) for some \(z_{0} \in \Omega\). Then \(\Omega\) contains \(a\) neighborhood \(V\) of \(z_{0}\) such that
\(\varphi\) is one-to-one in \(V\);
\(W=\varphi(V)\) is an open set;
if \(\psi: W \rightarrow V\) is defined by \(\psi(\varphi(z))=z\), then \(\psi \in H(W)\).
Thus \(\varphi: V \rightarrow W\) has a holomorphic inverse.
Proof: By the previous lemma applied to \(\varphi\) in place of \(f\), we conclude that \(\Omega\) contains a neighborhood \(V\) of \(z_{0}\) such that \[\left|\varphi\left(z_{1}\right)-\varphi\left(z_{2}\right)\right| \geq \frac{1}{2}\left|\varphi^{\prime}\left(z_{0}\right)\right|\left|z_{1}-z_{2}\right| \tag{1}\] if \(z_{1} \in V\) and \(z_{2} \in V\). Thus (a) holds, and \(\varphi^{\prime}(z) \neq 0\) for \(z\in V\).
To prove (b), pick \(a \in V\) and choose \(r>0\) so that \(\overline{D}(a, r) \subseteq V\).
By inequality (1) there exists \(c>0\) such that \[\big|\varphi\big(a+r e^{i \theta}\big)-\varphi(a)\big|>2 c \quad \text{ for } \quad \theta\in [-\pi,\pi]. \tag{2}\]
If \(\lambda \in D(\varphi(a) ; c)\), then \(|\lambda-\varphi(a)|<c\), hence (2) implies \[\min _{\theta\in [-\pi,\pi]}\big|\lambda-\varphi\big(a+r e^{i \theta}\big)\big|>c>|\lambda-\varphi(a)|. \tag{3}\]
By the previous corollary \(\lambda-\varphi\) must therefore have a zero in \(D(a ; r)\).
Thus \(\lambda=\varphi(z)\) for some \(z \in D(a ; r) \subseteq V\).
This proves that \(D(\varphi(a) ; c) \subseteq \varphi(V)\).
Hence \(\varphi(V)\) is open, since \(a\) was an arbitrary point of \(V\).
To prove (c), fix \(w_{1} \in W\).
Then \(\varphi\left(z_{1}\right)=w_{1}\) for a unique \(z_{1} \in V\) by property (b).
If \(w \in W\) and \(\psi(w)=z \in V\), we have
\[\frac{\psi(w)-\psi\left(w_{1}\right)}{w-w_{1}}=\frac{z-z_{1}}{\varphi(z)-\varphi\left(z_{1}\right)}.\]
By inequality (1) we deduce that \(z \rightarrow z_{1}\) when \(w \rightarrow w_{1}\).
Moreover, \(\phi'(z)\neq0\) if \(z\in V\), also by (1).
Hence \[\psi^{\prime}\left(w_{1}\right)=\lim_{w\to w_1}\frac{\psi(w)-\psi\left(w_{1}\right)}{w-w_{1}}=\lim_{z\to z_1}\frac{z-z_{1}}{\varphi(z)-\varphi\left(z_{1}\right)}=1 / \varphi^{\prime}\left(z_{1}\right).\]
Thus \(\psi \in H(W)\) as desired. $$\tag*{$\blacksquare$}$$
For \(m\in \mathbb N\), we denote the \(m^{\text{th}}\)-power function \(z \mapsto z^{m}\) by \(\pi_{m}\).
Each \(w \neq 0\) is \(\pi_{m}(z)\) for precisely \(m\) distinct values of \(z\): If \(w=r e^{i \theta}\) for some \(r>0\), then \[\pi_{m}(z)=w \quad \iff \quad z=r^{1 / m} e^{i(\theta+2 k \pi) / m} \quad \text{ for } \quad k=1, \ldots, m.\]
Note also that each \(\pi_{m}\) is an open mapping: If \(V\) is open and does not contain \(0\), then \(\pi_{m}(V)\) is open by the previous theorem. On the other hand, \(\pi_{m}(D(0, r))=D\left(0, r^{m}\right)\).
Compositions of open mappings are clearly open. In particular, \(\pi_{m} \circ \varphi\) is open, by the previous theorem, if \(\varphi^{\prime}\) has no zero.
Suppose \(\Omega\subseteq \mathbb C\) is open, \(f \in H(\Omega)\) and \(f\) is not constant, \(z_{0} \in \Omega\), and \(w_{0}=f\left(z_{0}\right)\). Let \(m\) be the order of the zero which the function \(f-w_{0}\) has at \(z_{0}\). Then there exists a neighborhood \(V \subseteq \Omega\) of \(z_{0}\), and there exists \(\varphi \in H(V)\), such that
\(f(z)=w_{0}+[\varphi(z)]^{m}\) for all \(z \in V\).
Moreover, \(\varphi^{\prime}\) has no zero in \(V\) and \(\varphi\) is an invertible mapping of \(V\) onto a disc \(D(0 ; r)\).
Remark. Thus \(f-w_{0}=\pi_{m} \circ \varphi\) in \(V\). It follows that \(f\) is an exactly \(m\)-to-1 mapping of \(V\setminus\left\{z_{0}\right\}\) onto \(D^{\prime}\left(w_{0} ; r^{m}\right)\), and that each \(w_{0} \in f(\Omega)\) is an interior point of \(f(\Omega)\). Hence \(f(\Omega)\) is open.
Proof: Without loss of generality we may assume that \(\Omega\) is a convex neighborhood of \(z_{0}\) which is so small that \(f(z) \neq w_{0}\) if \(z \in \Omega\setminus\left\{z_{0}\right\}\).
Then \[f(z)-w_{0}=\left(z-z_{0}\right)^{m} g(z) \quad \text{ for } \quad z \in \Omega\] for some \(g \in H(\Omega)\) which has no zero in \(\Omega\). Hence \(g^{\prime} / g \in H(\Omega)\).
By the Cauchy theorem \(g^{\prime} / g=h^{\prime}\) for some \(h \in H(\Omega)\).
The derivative of \(g \cdot \exp (-h)\) is \(0\) in \(\Omega\).
If \(h\) is modified by the addition of a suitable constant, it follows that \(g=\exp (h)\). Define \[\varphi(z)=\left(z-z_{0}\right) \exp \frac{h(z)}{m} \quad \text{ for } \quad z \in \Omega.\]
Then (a) holds, for all \(z \in \Omega\).
Also, \(\varphi\left(z_{0}\right)=0\) and \(\varphi^{\prime}\left(z_{0}\right) \neq 0\). The existence of an open set \(V\) that satisfies \((b)\) follows now from the previous theorem. $$\tag*{$\blacksquare$}$$
Suppose that \(\Omega\subseteq \mathbb C\) is open, \(f \in H(\Omega)\), and \(f\) is one-to-one in \(\Omega\). Then \(f^{\prime}(z) \neq 0\) for every \(z \in \Omega\), and the inverse of \(f\) is holomorphic.
Proof: If \(f^{\prime}\left(z_{0}\right)\) were \(0\) for some \(z_{0} \in \Omega\), the hypotheses of the previous theorem would hold with some \(m>1\), so that \(f\) would be \(m\)-to-\(1\) in some deleted neighborhood of \(z_{0}\), which is impossible, since \(f\) is one-to-one. Thus \(f^{\prime}\left(z\right)\neq 0\) for all \(z\in \Omega\). Now apply part (c) of the last but one theorem. This completes the proof of the theorem. $$\tag*{$\blacksquare$}$$
Remark. We observe that the converse of the inverse mapping theorem is false: If \(f(z)=e^{z}\), then \(f^{\prime}(z) \neq 0\) for every \(z\in\mathbb C\), but \(f\) is not one-to-one in the whole complex plane.
The exponential function \(S_{\alpha}\ni z\mapsto e^z\in \mathbb C\setminus\{0\}\) when restricted to the strip \(S_{\alpha}=\{x+i y: \alpha \leq y<\alpha+2 \pi\}\) is a one-to-one analytic map of this strip onto \(\mathbb C\setminus\{0\}\) the nonzero complex numbers.
We take \(\log _{\alpha}\) to be the inverse of the exponential function restricted to the strip \(S_{\alpha}= \{x+i y: \alpha \leq y<\alpha+2 \pi\}\).
We define \(\arg _{\alpha}\) to the imaginary part of \(\log _{\alpha}\).
Consequently, \(\log _{\alpha}(\exp z)=z\) for each \(z \in S_{\alpha}\), and \(\exp \left(\log _{\alpha} z\right)=z\) for all \(z \in \mathbb{C} \backslash\{0\}\).
The principal branches of the logarithm and argument functions, to be denoted by \(\operatorname{Log}\) and \(\operatorname{Arg}\), are obtained by taking \(\alpha=-\pi\).
Thus, \(\operatorname{Log} =\log _{-\pi}\) and \(\operatorname{Arg}=\arg _{-\pi}\).
If \(z \neq 0\), then \(\log _{\alpha}(z)=\log |z|+i \arg _{\alpha}(z)\), and \(\arg _{\alpha}(z)\) is the unique number in \([\alpha, \alpha+2 \pi)\) such that \[z /|z|=e^{i \arg _{\alpha}(z)}.\] In other words, the unique argument of \(z\) in \([\alpha, \alpha+2 \pi)\).
Let \(R_{\alpha}=\left\{r e^{i \alpha}: r \geq 0\right\}\). The functions \(\log _{\alpha}\) and \(\arg _{\alpha}\) are continuous at each point of the "slit" complex plane \(\mathbb{C} \backslash R_{\alpha}\), and discontinuous at each point of \(R_{\alpha}\).
The function \(\log _{\alpha}\) is analytic on \(\mathbb{C} \backslash R_{\alpha}\), and its derivative is given by \(\log _{\alpha}^{\prime}(z)=1 / z\).
Proof:
If \(w=\log _{\alpha}(z)\) with \(z \neq 0\), then \(e^{w}=z\), hence \[|z|=e^{\operatorname{Re} w}, \quad \text{ and } \quad z /|z|=e^{i \operatorname{Im} w}.\] Thus \(\operatorname{Re} w=\log |z|\), and \(\operatorname{Im} w\) is an argument of \(z /|z|\). Since \(\operatorname{Im} w\) is restricted to \([\alpha, \alpha+2 \pi)\) by definition of \(\log _{\alpha}\), it follows that \(\operatorname{Im} w\) is the unique argument for \(z\) that lies in the interval \([\alpha, \alpha+2 \pi)\).
By (a), it suffices to consider \(\arg _{\alpha}\). If \(z_{0} \in \mathbb{C} \backslash R_{\alpha}\) and \((z_{n})_{n\in\mathbb N}\) is a sequence converging to \(z_{0}\), then \(\arg _{\alpha}\left(z_{n}\right)\) must converge to \(\arg _{\alpha}\left(z_{0}\right)\). On the other hand, if \(z_{0} \in R_{\alpha} \backslash\{0\}\), there is a sequence \((z_{n})_{n\in\mathbb N}\) converging to \(z_{0}\) so that \[\lim_{n\to\infty}\arg _{\alpha}\left(z_{n}\right)= \alpha+2 \pi \neq \arg _{\alpha}\left(z_{0}\right)=\alpha.\]
Recall from Lecture 2 the following theorem:
Let \(g\) be analytic on the open set \(\Omega_{1}\), and let \(f\) be a continuous complex-valued function on the open set \(\Omega\). Assume that
\(f(\Omega) \subseteq \Omega_{1}\),
\(g^{\prime}\) is never 0 ,
\(g(f(z))=z\) for all \(z \in \Omega\) (thus \(f\) is one-to-one).
Then \(f\) is analytic on \(\Omega\) and \(f^{\prime}=1 /\left(g^{\prime} \circ f\right)\).
By this theorem with \(g=\exp\) , \(\Omega_{1}=\mathbb{C}\), \(f=\log _{\alpha}\), and \(\Omega=\mathbb{C} \backslash R_{\alpha}\) and the fact that \(\exp\) is its own derivative we obtain that \[(\log _{\alpha} z)'=\frac{1}{z}.\] This completes the proof of the theorem.$$\tag*{$\blacksquare$}$$
Let \(S\) be a subset of \(\mathbb{C}\) (or more generally any metric space), and let \(f: S \rightarrow \mathbb{C} \backslash\{0\}\) be continuous.
A function \(g: S \rightarrow \mathbb{C}\) is a continuous logarithm of \(f\) if \(g\) is continuous on \(S\) and \(f(s)=e^{g(s)}\) for all \(s \in S\).
A function \(\theta: S \rightarrow \mathbb{R}\) is a continuous argument of \(f\) if \(\theta\) is continuous on \(S\) and \(f(s)=|f(s)| e^{i \theta(s)}\) for all \(s \in S\).
Examples
If \(S=[0,2 \pi]\) and \(f(s)=e^{i s}\), then \(f\) has a continuous argument on \(S\), namely \(\theta(s)=s+2 k \pi\) for any fixed integer \(k\).
If \(f\) is a continuous mapping of \(S\) into \(\mathbb{C} \backslash R_{\alpha}\) for some \(\alpha\in \mathbb R\), then \(f\) has a continuous argument, namely \(\theta(s)=\arg _{\alpha}(f(s))\).
If \(S=\{z:|z|=1\}\) and \(f(z)=z\), then \(f\) does not have a continuous argument on \(S\).
Let \(f: S \rightarrow \mathbb{C}\) be continuous.
If \(g\) is a continuous logarithm of \(f\), then \(\operatorname{Im} g\) is a continuous argument of \(f\).
If \(\theta\) is a continuous argument of \(f\), then \(\log |f|+i \theta\) is a continuous logarithm of \(f\). Thus \(f\) has a continuous logarithm iff \(f\) has a continuous argument.
Assume that \(S\) is connected, and \(f\) has continuous logarithms \(g_{1}\) and \(g_{2}\), and continuous arguments \(\theta_{1}\) and \(\theta_{2}\). Then there are integers \(k\) and \(l\) such that \(g_{1}(s)-g_{2}(s)=2 \pi i k\) and \(\theta_{1}(s)-\theta_{2}(s)=2 \pi l\) for all \(s \in S\). Thus \(g_{1}-g_{2}\) and \(\theta_{1}-\theta_{2}\) are constant on \(S\).
If \(S\) is connected and \(s, t \in S\), then \[g(s)-g(t)=\log |f(s)|-\log |f(t)|+i(\theta(s)-\theta(t))\] for all continuous logarithms \(g\) and all continuous arguments \(\theta\) of \(f\).
Proof:
If \(f(s)=e^{g(s)}\), then \(|f(s)|=e^{\operatorname{Re} g(s)}\), hence \(f(s) /|f(s)|=e^{i \operatorname{Im} g(s)}\) as required.
If \(f(s)=|f(s)| e^{i \theta(s)}\), then \(f(s)=e^{\log |f(s)|+i \theta(s)}\), so \(\log |f|+i \theta\) is a continuous logarithm of \(f\).
We have \(f(s)=e^{g_{1}(s)}=e^{g_{2}(s)}\), hence \(e^{g_{1}(s)-g_{2}(s)}=1\), for all \(s \in S\). Hence \(g_{1}(s)-g_{2}(s)=2 \pi i k(s)\) for some integer-valued function \(S\ni s\mapsto k(s)\in\mathbb Z\). Since \(g_{1}\) and \(g_{2}\) are continuous on \(S\), so is \(k\). But \(S\) is connected, so \(k\) is a constant function. A similar proof applies to any pair of continuous arguments of \(f\).
If \(\theta\) is a continuous argument of \(f\), then \(\log |f|+i \theta\) is a continuous logarithm of \(f\) by part (b). Thus if \(g\) is any continuous logarithm of \(f\), then \(g=\log |f|+i \theta+2 \pi i k\) by (c).
The result follows.$$\tag*{$\blacksquare$}$$
Let \(\gamma:[a, b] \rightarrow \mathbb{C} \backslash\{0\}\) be a continuous curve and \(0 \notin \gamma^{*}\). Then \(\gamma\) has a continuous argument, and consequently a continuous logarithm.
Proof: Let \(\varepsilon=\inf \{|\gamma(t)|: t \in[a, b]\}\) be the distance from \(0\) to \(\gamma^{*}\).
Then \(\varepsilon>0\) because \(0 \notin \gamma^{*}\) and \(\gamma^{*}\) is a closed set.
By the uniform continuity of \(\gamma\) on \([a, b]\), there is a partition \(a=t_{0}<t_{1}<\cdots<t_{n}=b\) of \([a, b]\) such that if \(1 \leq j \leq n\) and \(t \in\left[t_{j-1}, t_{j}\right]\), then \(\gamma(t) \in D\left(\gamma\left(t_{j}\right), \varepsilon\right)\).
Since \(\gamma: [t_{j-1}, t_{j}]\to D\left(\gamma\left(t_{j}\right),\varepsilon\right)\) is a continuous function and \(D\left(\gamma\left(t_{j}\right),\varepsilon\right)\subseteq \mathbb C\setminus R_{\alpha_j}\) for some \(\alpha_j\in[0, 2\pi]\), then each \(\gamma_{|[t_{j-1}, t_{j}]}\) has a continuous argument \(\theta_j(t)=\arg _{\alpha_j}(\gamma_{|[t_{j-1}, t_{j}]}(t))\)
Since \(\theta_{j}\left(t_{j}\right)\) and \(\theta_{j+1}\left(t_{j}\right)\) differ by an integer multiple of \(2 \pi\), we may (if necessary) redefine \(\theta_{j+1}\) on \(\left[t_{j}, t_{j+1}\right]\) so that the relation \(\theta_{j} \cup \theta_{j+1}\) is a continuous argument of \(\gamma\) on \(\left[t_{j-1}, t_{j+1}\right]\). Proceeding in this way, we obtain a continuous argument of \(\gamma\) on the entire interval \([a, b]\). $$\tag*{$\blacksquare$}$$
Let \(\gamma:[a, b] \rightarrow \mathbb{C}\) be a closed curve. Fix \(z_{0} \notin \gamma^{*}\), and let \(\theta\) be a continuous argument of \(\gamma-z_{0}\), (\(\theta\) exists by the previous theorem). Then \(\theta(b)-\theta(a)\) is an integer multiple of \(2 \pi\). Furthermore, if \(\theta_{1}\) is another continuous argument of \(\gamma-z_{0}\), then \(\theta_{1}(b)-\theta_{1}(a)=\theta(b)-\theta(a)\).
Proof: We know that \(\left(\gamma(t)-z_{0}\right) /\left|\gamma(t)-z_{0}\right|=e^{i \theta(t)}\),for \(t\in[a,b]\).
Since \(\gamma\) is a closed curve, \(\gamma(a)=\gamma(b)\), hence \[1=\frac{\gamma(b)-z_{0}}{\left|\gamma(b)-z_{0}\right|} \cdot \frac{\left|\gamma(a)-z_{0}\right|}{\gamma(a)-z_{0}}=e^{i(\theta(b)-\theta(a))}.\]
Consequently, \(\theta(b)-\theta(a)\) is an integer multiple of \(2 \pi\). If \(\theta_{1}\) is another continuous argument of \(\gamma-z_{0}\), then we have that \(\theta_{1}-\theta=2 \pi l\) for some \(l\in\mathbb Z\). Thus \(\theta_{1}(b)=\theta(b)+2 \pi l\) and \(\theta_{1}(a)=\theta(a)+2 \pi l\), so \(\theta_{1}(b)-\theta_{1}(a)=\theta(b)-\theta(a)\). $$\tag*{$\blacksquare$}$$
Let \(\gamma:[a, b] \rightarrow \mathbb{C}\) be a closed curve. If \(z_{0} \notin \gamma^{*}\), let \(\theta_{z_{0}}\) be a continuous argument of \(\gamma-z_{0}\). The winding number of \(z_{0}\) with respect to \(\gamma\), is defined by \[W\left(\gamma, z_{0}\right)=\frac{\theta_{z_{0}}(b)-\theta_{z_{0}}(a)}{2 \pi}.\]
Remarks.
By the previous theorem, \(W\left(\gamma, z_{0}\right)\) is well-defined, that is, \(W\left(\gamma, z_{0}\right)\) does not depend on the particular continuous argument chosen.
Intuitively, \(W\left(\gamma, z_{0}\right)\) is the net number of revolutions of \(\gamma(t)\) about the point \(z_{0}\) when \(t\in[a, b]\).
Note that by the above definition, for any complex number \(w\) we have \(W\left(\gamma, z_{0}\right)=W\left(\gamma+w, z_{0}+w\right)\).
Let \(\gamma:[a, b]\to \mathbb C\) be a closed path, and \(z_{0}\) a point not belonging to \(\gamma^{*}\). Then \[W\left(\gamma, z_{0}\right)=\frac{1}{2 \pi i} \int_{\gamma} \frac{1}{z-z_{0}} d z=\operatorname{Ind}_{\gamma}(z_0).\] In other words, the winding number is the index function. More generally, if \(f\) is analytic on an open set \(\Omega\) containing \(\gamma^{*}\), and \(z_{0} \notin(f \circ \gamma)^{*}\), then \[W\left(f \circ \gamma, z_{0}\right)=\frac{1}{2 \pi i} \int_{\gamma} \frac{f^{\prime}(z)}{f(z)-z_{0}} d z.\]
Proof: Let \(\varepsilon\) be the distance from \(z_{0}\) to \(\gamma^{*}\).
Since \(\gamma\) is uniformly continuous on \([a, b]\), then there is a partition \(a=t_{0}<t_{1}<\cdots<t_{n}=b\) so that \[\gamma(t) \in D\left(\gamma\left(t_{j}\right), \varepsilon\right) \quad \text{ for } \quad t\in [t_{j-1},t_{j}].\]
By definition of \(\varepsilon\) we have \(z_{0} \notin D\left(\gamma\left(t_{j}\right), \varepsilon\right)\) for each \(1\le j\le n\).
Consequently, the holomorphic function \(h(z)= z-z_{0}\), when restricted to \(D\left(\gamma\left(t_{j}\right), \varepsilon\right)\) is nonvanishing.
Therefore, \(\frac{h'(z)}{h(z)}=\frac{1}{\left(z-z_{0}\right)}\) is holomorphic in \(D\left(\gamma\left(t_{j}\right), \varepsilon\right)\).
By the Cauchy theorem \(h'/h\) has a primitive \(g_{j}\) in \(D\left(\gamma\left(t_{j}\right), \varepsilon\right)\).
Therefore \(g_{j}^{\prime}(z)=1 /\left(z-z_{0}\right)\) for all \(z \in D\left(\gamma\left(t_{j}\right), \varepsilon\right)\).
The path \(\gamma\) restricted to \(\left[t_{j-1}, t_{j}\right]\) lies in the disk \(D\left(\gamma\left(t_{j}\right), \varepsilon\right)\), hence \[\int_{\gamma_{|[t_{j-1}, t_{j}]}} \frac{1}{z-z_{0}} dz =g_{j}\left(\gamma\left(t_{j}\right)\right)-g_{j}\left(\gamma\left(t_{j-1}\right)\right).\]
Thus \[\int_{\gamma} \frac{1}{z-z_{0}} d z=\sum_{j=1}^{n}\left[g_{j}\left(\gamma\left(t_{j}\right)\right)-g_{j}\left(\gamma\left(t_{j-1}\right)\right)\right].\]
Observe that the function \(\exp(-g_j)h\) has a vanishing derivative on \(D\left(\gamma\left(t_{j}\right), \varepsilon\right)\). If \(g_j\) is modified by the addition of a suitable constant, it follows that \(h=\exp(g_j)\). If \(\theta_{j}=\operatorname{Im} g_{j}\), then \(\theta_{j}\) is a continuous argument of \(h(z)= z-z_{0}\) on \(D\left(\gamma\left(t_{j}\right), \varepsilon\right)\).
Since \(g_j(z)=\log|z|+i\theta_j(z)\), then \[\int_{\gamma} \frac{1}{z-z_{0}} d z=i \sum_{j=1}^{n}\left[\theta_{j}\left(\gamma\left(t_{j}\right)\right)-\theta_{j}\left(\gamma\left(t_{j-1}\right)\right)\right].\]
If \(\theta\) is any continuous argument of \(\gamma-z_{0}\), then \(\theta_{|[t_{j-1}, t_{j}]}\) is a continuous argument of \((\gamma-z_{0})_{|[t_{j-1}, t_{j}]}\). But so is \(\theta_{j} \circ \gamma_{|[t_{j-1}, t_{j}]}\), hence \[\theta_{j}\left(\gamma\left(t_{j}\right)\right)-\theta_{j}\left(\gamma\left(t_{j-1}\right)\right)=\theta\left(t_{j}\right)-\theta\left(t_{j-1}\right),\] since two continuous arguments differs by a multiple of \(2\pi\).
Therefore, \[\begin{aligned} \int_{\gamma} \frac{1}{z-z_{0}} d z&=i\sum_{j=1}^{n}\left[\theta\left(t_{j}\right)-\theta\left(t_{j-1}\right)\right]\\ &=i(\theta(b)-\theta(a))\\ &=2 \pi iW\left(\gamma, z_{0}\right) \end{aligned}\] completing the proof of the first part of the theorem.
Applying this result to the path \(f \circ \gamma\), we get the second statement. Specifically, if \(z_{0} \notin(f \circ \gamma)^{*}\), then \[\begin{aligned} W\left(f \circ \gamma, z_{0}\right)&=\frac{1}{2 \pi i} \int_{f \circ \gamma} \frac{1}{z-z_{0}} d z\\ &=\frac{1}{2 \pi i} \int_{\gamma} \frac{f^{\prime}(z)}{f(z)-z_{0}} d z. \qquad \end{aligned}\tag*{$\blacksquare$}\]