Recall that
Let \(\Omega\subseteq \mathbb C\) be a convex open set and \(p \in \Omega\). Let \(f\) be continuous in \(\Omega\) and holomorphic in \(\Omega \backslash\{p\}\). Then \(f\) has a primitive in \(\Omega\) and \[\int_{\gamma} f(z) d z=0\] for any closed path \(\gamma\) in \(\Omega\).
Let \(\gamma\) be a closed path in an open convex set \(\Omega\subseteq \mathbb C\) and \(f \in H(\Omega)\). Then for any \(a \in \Omega\) and \(a \notin \gamma^{*}\), we have \[\frac{1}{2 \pi i} \int_{\gamma} \frac{f(z)}{z-a} d z=\operatorname{Ind}_{\gamma}(a) f(a).\]
Proof: We define for \(z \in \Omega\) the following function \[g(z)= \begin{cases}\frac{f(z)-f(a)}{z-a} & \text { if } z \neq a \\ f^{\prime}(a) & \text { if } z=a\end{cases}\]
We observe that \(g\) is continuous in \(\Omega\) and \(g \in H(\Omega \backslash\{a\})\).
Now we apply the Cauchy theorem for convex sets (with \(f=g\) and \(p=a\)), to conclude that \[\int_{\gamma} g(z) d z=0.\]
Then, since \(a \notin \gamma^{*}\), we have \[\frac{1}{2 \pi i} \int_{\gamma} \frac{f(z)}{z-a} d z=\frac{f(a)}{2 \pi i} \int_{\gamma} \frac{d z}{z-a}=\operatorname{Ind}_{\gamma}(a) f(a).\qquad \tag*{$\blacksquare$}\]
If \(f\) is holomorphic in an open set \(\Omega\subseteq \mathbb C\), then \(f\) has infinitely many complex derivatives in \(\Omega\). In other words,
If \(f\in H(\Omega)\), then \(f^{(n)}\in H(\Omega)\) for any \(n\in\mathbb N\).
Moreover, if \(C \subset \Omega\) is a circle whose interior is also contained in \(\Omega\), then \[f^{(n)}(z)=\frac{n!}{2 \pi i} \int_{C} \frac{f(\zeta) d \zeta}{(\zeta-z)^{n+1}}\] for all \(z\) in the interior of \(C\).
Proof: The proof is by induction on \(n\ge0\), the case \(n=0\) being simply the Cauchy integral formula.
Suppose that \(f\) has up to \(n-1\) complex derivatives and that \[f^{(n-1)}(z)=\frac{(n-1)!}{2 \pi i} \int_{C} \frac{f(\zeta)}{(\zeta-z)^{n}d \zeta}.\]
Now for \(h\) small, the difference quotient for \(f^{(n-1)}\) takes the form \[\begin{aligned} & \frac{f^{(n-1)}(z+h)-f^{(n-1)}(z)}{h}= \\ & \quad \frac{(n-1)!}{2 \pi i} \int_{C} f(\zeta) \frac{1}{h}\left[\frac{1}{(\zeta-z-h)^{n}}-\frac{1}{(\zeta-z)^{n}}\right] d \zeta \end{aligned}\]
We now recall that \[A^{n}-B^{n}=(A-B)\left[A^{n-1}+A^{n-2} B+\cdots+A B^{n-2}+B^{n-1}\right] .\]
With \(A=1 /(\zeta-z-h)\) and \(B=1 /(\zeta-z)\), we see that the term in brackets under the integral is equal to \[\frac{h}{(\zeta-z-h)(\zeta-z)}\left[A^{n-1}+A^{n-2} B+\cdots+A B^{n-2}+B^{n-1}\right].\]
Observe that if \(h\) is small, then \(z+h\) and \(z\) stay at a finite distance from the boundary circle \(C\), so in the limit as \(h\) tends to \(0\), we find that the quotient converges to \[\begin{aligned} &\frac{(n-1)!}{2 \pi i} \int_{C} f(\zeta)\left[\frac{1}{(\zeta-z)^{2}}\right]\left[\frac{n}{(\zeta-z)^{n-1}}\right] d \zeta\\ &\qquad =\frac{n!}{2 \pi i} \int_{C} \frac{f(\zeta)}{(\zeta-z)^{n+1}} d \zeta, \end{aligned}\] which completes the induction argument and proves the theorem. $$\tag*{$\blacksquare$}$$
If \(f\) is holomorphic in an open set \(\Omega\subseteq \mathbb C\) that contains a closed disc \(\overline{D}(z_0, R)\) centered at \(z_{0}\) and of radius \(R\), then \[\big|f^{(n)}\left(z_{0}\right)\big| \leq \frac{n!\|f\|_{L^\infty(C)}}{R^{n}},\] where \(\|f\|_{L^\infty(C)}=\sup _{z \in C}|f(z)|\) denotes the supremum of \(|f|\) on the boundary circle \(C=\partial \overline{D}(z_0, R)\).
Proof: Applying the Cauchy integral formula for \(f^{(n)}\left(z_{0}\right)\), we obtain \[\begin{aligned} \big|f^{(n)}\left(z_{0}\right)\big| & =\left|\frac{n!}{2 \pi i} \int_{C} \frac{f(\zeta)d \zeta}{\left(\zeta-z_{0}\right)^{n+1}} \right| \\ & =\frac{n!}{2 \pi}\bigg|\int_{0}^{2 \pi} \frac{f\left(z_{0}+R e^{i \theta}\right)}{\left(R e^{i \theta}\right)^{n+1}} R i e^{i \theta} d \theta\bigg| \leq \frac{n!}{2 \pi} \frac{\|f\|_{L^\infty(C)}}{R^{n}} 2 \pi.\qquad \end{aligned}\tag*{$\blacksquare$}\]
Suppose \(f\) is holomorphic in an open set \(\Omega\subseteq \mathbb C\). If \(D\) is a disc centered at \(z_{0}\) and whose closure is contained in \(\Omega\), then \(f\) has a power series expansion at \(z_{0}\), given by \[f(z)=\sum_{n=0}^{\infty} a_{n}\left(z-z_{0}\right)^{n}\] for all \(z \in D\), and the coefficients are given by \[a_{n}=\frac{f^{(n)}\left(z_{0}\right)}{n!} \quad \text { for all } \quad n \geq 0.\]
Observe that since power series define indefinitely (complex) differentiable functions, the theorem gives another proof that a holomorphic function is automatically indefinitely differentiable.
Proof: Fix \(z \in D\). By the Cauchy integral formula, we have \[f(z)=\frac{1}{2 \pi i} \int_{C} \frac{f(\zeta)}{\zeta-z} d \zeta,\] where \(C\) denotes the boundary of the disc \(D\) and \(z \in D\).
The idea is to write \[\frac{1}{\zeta-z}=\frac{1}{\zeta-z_{0}-\left(z-z_{0}\right)}=\frac{1}{\zeta-z_{0}} \frac{1}{1-\left(\frac{z-z_{0}}{\zeta-z_{0}}\right)},\] and use the geometric series expansion.
Since \(\zeta \in C\) and \(z \in D\) is fixed, there exists \(0<r<1\) such that \[\left|\frac{z-z_{0}}{\zeta-z_{0}}\right|<r.\]
Therefore \[\frac{1}{1-\left(\frac{z-z_{0}}{\zeta-z_{0}}\right)}=\sum_{n=0}^{\infty}\left(\frac{z-z_{0}}{\zeta-z_{0}}\right)^{n},\] where the series converges uniformly for \(\zeta \in C\).
This allows us to interchange the infinite sum with the integral when we combine the last three identities, thereby obtaining \[f(z)=\sum_{n=0}^{\infty}\left(\frac{1}{2 \pi i} \int_{C} \frac{f(\zeta)d \zeta}{\left(\zeta-z_{0}\right)^{n+1}} \right) \cdot\left(z-z_{0}\right)^{n}.\]
This proves the power series expansion; further the use of the Cauchy integral formulas for the derivatives (or simply differentiation of the series) proves the formula for \(a_{n}\). $$\tag*{$\blacksquare$}$$
Remarks.
Another important observation is that the power series expansion of \(f\) centered at \(z_{0}\) converges in any disc, no matter how large, as long as its closure is contained in \(\Omega\).
In particular, if \(f\) is entire (that is, holomorphic on all of \(\mathbb{C}\) ), the previous theorem implies that \(f\) has a power series expansion around \(0\), say \(f(z)=\sum_{n=0}^{\infty} a_{n} z^{n}\), that converges in all of \(\mathbb{C}\).
Let \(\Omega\subseteq \mathbb C\) be an open set. Then the following statements are exuivalent.
\(f\in H(\Omega)\).
\(f^{(n)}\in H(\Omega)\) for every \(n\ge 0\).
If \(D\) is a disc centered at \(z_{0}\) and whose closure is contained in \(\Omega\), then \(f\) has a power series expansion at \(z_{0}\), given by \[f(z)=\sum_{n=0}^{\infty} a_{n}\left(z-z_{0}\right)^{n}\] for all \(z \in D\), and the coefficients are given by \[a_{n}=\frac{f^{(n)}\left(z_{0}\right)}{n!} \quad \text { for all } \quad n \geq 0.\]
Let \(0 \leq s_{1}<s_{2} \leq+\infty\) and \(z_{0} \in \mathbb{C}\) be given.
An open annulus centered at \(z_0\) with radii \(s_1\), and \(s_2\) is defined by \[A\left(z_{0}, s_{1}, s_{2}\right)=\left\{z\in \mathbb C: s_{1}<\left|z-z_{0}\right|<s_{2}\right\}.\]
A closed annulus centered at \(z_0\) with radii \(s_1\), and \(s_2\) is defined by \[\overline{A}\left(z_{0}, s_{1}, s_{2}\right)=\left\{z\in \mathbb C: s_{1}\le \left|z-z_{0}\right|\le s_{2}\right\}.\]
Note that \[\begin{aligned} A\left(z_{0}, s_{1}, s_{2}\right) &= D(z_0, s_2)\setminus \overline{D}(z_0, s_1),\\ \overline{A}\left(z_{0}, s_{1}, s_{2}\right) &= \overline{D}(z_0, s_2)\setminus D(z_0, s_1),\\ \overline{A}\left(z_{0}, s_{1}, s_{2}\right) &=A\left(z_{0}, s_{1}, s_{2}\right)\cup C(z_0, s_1)\cup C(z_0, s_2). \end{aligned}\]
Let \(z_0\in\mathbb C\). If two series \[\sum_{n\ge0}a_n(z-z_0)^n, \quad \text{ and } \quad \sum_{n<0}a_n(z-z_0)^n\tag{*}\] converge for some \(z\in\mathbb C\), then their sum \[\sum_{n\in \mathbb Z}a_n(z-z_0)^n \tag{**}\] also converges at \(z\in\mathbb C\).
The sum in (**) is called the Laurent series centered at \(z_0\).
The sum (**) is said to diverge if at least one of the series in (*) diverges.
Let \(z_0\in\mathbb C\) \[f(z)=\sum_{n\in \mathbb Z}a_n(z-z_0)^n\] be the Laurent series, and define \[r=\lim \sup _{n \rightarrow \infty}\sqrt[n]{|a_{-n}|}, \quad\text{ and } \quad R=\Big[\lim \sup _{n \rightarrow \infty}\sqrt[n]{|a_{n}|}\Big]^{-1}.\] If \(r<R\), then the Laurent series \(f(z)\) converges absolutely on the annulus \[A(z_0, r, R)=\{z\in\mathbb C: r<|z-z_0|<R\}\] and uniformly on its compact subsets. Therefore, the Laurent series \(f(z)\) defines a holomorphic function on \(A(z_0, r, R)\).
Proof: We will analyze the series from (*) and (**) separately.
Note that the first series \[g(z)=\sum_{n\ge0}a_n(z-z_0)^n\] is the power series centered at \(z_0\in\mathbb C\).
Its radius of convergence is given by the formula \[R=\Big[\lim \sup _{n \rightarrow \infty}\sqrt[n]{|a_{n}|}\Big]^{-1}.\]
We know that the series converges absolutely on \(D\left(z_{0}, R\right)\), and uniformly on its compact subsets, and diverges on \(\overline{D}\left(z_{0}, R\right)^c\).
Here we adopt the convention that \(1 / 0=\infty, 1 / \infty=0\).
Consequently, \(g(z)\) is a holomorphic function on \(D\left(z_{0}, R\right)\).
If we set \(w=\frac{1}{z-z_0}\) then we obtain a new power series \[h(z)=\sum_{n<0}a_n(z-z_0)^n=\sum_{n\ge1}a_{-n}w^n,\] with the corresponding radius of convergence \(\rho=r^{-1}\).
Therefore, the series (of variable \(w\)) converges absolutely on \(D\left(0, \rho\right)\), and uniformly on its compact subsets, and diverges on \(\overline{D}\left(0, \rho\right)^c\).
But this is equivalent to say that the original series (of variable \(z\)) converges absolutely on \(\overline{D}\left(z_0, r\right)^c\), and uniformly on its compact subsets, and diverges on \(D\left(z_0, r\right)\).
Consequently, \(h(z)\) is a holomorphic function on \(\overline{D}\left(z_0, r\right)^c\).
Since \(r<R\) and \(\overline{D}\left(z_0, r\right)^c\cap D\left(z_{0}, R\right)=A(z_0, r, R)\) we conclude that \(f(z)=g(z)+h(z)\) is holomorphic on \(A(z_0, r, R)\) and desired. $$\tag*{$\blacksquare$}$$
Let \(f\) be analytic on an open set \(\Omega\) containing the annulus \(\overline{A}\left(z_{0}, r_{1}, r_{2}\right)\) for \(0<r_{1}<r_{2}<\infty\), and let \(\gamma_{1}\) and \(\gamma_{2}\) denote the positively oriented inner and outer boundaries of the annulus. Then for \(z\in A\left(z_{0}, r_{1}, r_{2}\right)\), we have \[f(z)=\frac{1}{2 \pi i} \int_{\gamma_{2}} \frac{f(w)}{w-z} d w-\frac{1}{2 \pi i} \int_{\gamma_{1}} \frac{f(w)}{w-z} d w\]
Proof: We can assume that \(\overline{A}\left(z_{0}, r_{1}, r_{2}\right)\subset A\left(z_{0}, s_{1}, s_{2}\right)\subseteq \Omega\) for some \(0<s_1<r_1<r_2<s_2\).
Note that \(A\left(z_{0}, s_{1}, s_{2}\right)\) is the region in which the Cauchy integral formula is available. Why?
Let \(\Gamma=I_1\cup C_2\cup I_2\cup C_1\subset A\left(z_{0}, s_{1}, s_{2}\right)\) be a path consisting of the following curves:
\(I_1\) is the positively oriented interval \([r_1, r_2]\subset \mathbb R\) and \(I_2=-I_1=[r_2, r_1]\);
\(C_1=-\gamma_1\), where \(\gamma_1\) is the positively oriented circle centered at \(z_0\) and radius \(r_1\); and \(C_2=\gamma_2\) is the positively oriented circle centered at \(z_0\) and radius \(r_2\).
Then by the Cauchy integral formula for \(A\left(z_{0}, s_{1}, s_{2}\right)\), we obtain \[\begin{aligned} f(z)&=\frac{1}{2 \pi i} \int_{\Gamma} \frac{f(w)}{w-z} d w\\ &=\frac{1}{2 \pi i} \int_{C_2} \frac{f(w)}{w-z} d w+\frac{1}{2 \pi i} \int_{C_1} \frac{f(w)}{w-z} d w\\ &=\frac{1}{2 \pi i} \int_{\gamma_{2}} \frac{f(w)}{w-z} d w-\frac{1}{2 \pi i} \int_{\gamma_{1}} \frac{f(w)}{w-z} d w, \end{aligned}\] since \[\frac{1}{2 \pi i} \int_{I_1} \frac{f(w)}{w-z} d w+ \frac{1}{2 \pi i} \int_{I_2} \frac{f(w)}{w-z} d w=0.\qquad \tag*{$\blacksquare$}\]
If \(f\) is holomorphic on \(\Omega=A\left(z_{0}, s_{1}, s_{2}\right)\), then there is a unique two-tailed sequence \((a_{n})_{n\in\mathbb Z}\subset \mathbb C\) such that \[f(z)=\sum_{n=-\infty}^{\infty} a_{n}\left(z-z_{0}\right)^{n}, \quad \text{ for } \quad z \in \Omega.\]
In fact, if \(r\) is such that \(s_{1}<r<s_{2}\), then the coefficients \(a_{n}\) are given by \[a_{n}=\frac{1}{2 \pi i} \int_{C\left(z_{0}, r\right)} \frac{f(w) d w}{\left(w-z_{0}\right)^{n+1}}, \quad \text{ for } \quad n\in\mathbb Z.\]
Also, the above series converges absolutely on \(\Omega\) and uniformly on compact subsets of \(\Omega\).
Proof: It suffices to prove this theorem for any compact subset of \(\Omega\). But any compact subset of \(\Omega\) is contained in closed annulus \[\overline{A}(z_0, \rho_1, \rho_2)\] for some \(\rho_{1}\) and \(\rho_{2}\) with \(s_{1}<\rho_{1}<\rho_{2}<s_{2}\).
We choose \(r_{1}\) and \(r_{2}\) such that \(s_{1}<r_{1}<\rho_1<\rho_2<r_{2}<s_{2}\). Then \[\overline{A}(z_0, \rho_1, \rho_2) \subset A(z_0, r_1, r_2)\subset \Omega=A\left(z_{0}, s_{1}, s_{2}\right).\]
By the previous theorem, for any \(z\in A(z_0, r_1, r_2)\), we have \[f(z)=\frac{1}{2 \pi i} \int_{C(z_0, r_2)} \frac{f(w)}{w-z} d w -\frac{1}{2 \pi i} \int_{C(z_0, r_1)} \frac{f(w)}{w-z} d w.\]
Consider the Cauchy type integral \[\frac{1}{2 \pi i} \int_{C\left(z_{0}, r_{2}\right)} \frac{f(w)}{w-z} d w \quad \text{ for } \quad z \in D\left(z_{0}, r_{2}\right).\]
Then proceeding as before, we obtain \[\frac{1}{2 \pi i} \int_{C\left(z_{0}, r_{2}\right)} \frac{f(w)}{w-z} d w=\sum_{n=0}^{\infty} a_{n}\left(z-z_{0}\right)^{n},\] where \[a_{n}=\frac{1}{2 \pi i} \int_{C\left(z_{0}, r_{2}\right)} \frac{f(w)d w}{\left(w-z_{0}\right)^{n+1}}.\]
The series converges absolutely on \(D\left(z_{0}, r_{2}\right)\), and uniformly on compact subsets of \(D\left(z_{0}, r_2\right)\).
Next, consider the Cauchy type integral \[-\frac{1}{2 \pi i} \int_{C\left(z_{0}, r_{1}\right)} \frac{f(w)}{w-z} d w\quad \text{ for } \quad z\in \overline{D}(z_0, r_1)^c.\]
This can be written as \[\begin{aligned} &\frac{1}{2 \pi i} \int_{C\left(z_{0}, r_{1}\right)} \frac{f(w) d w}{\left(z-z_{0}\right)\left[1-\frac{w-z_{0}}{z-z_{0}}\right]}\\ &\qquad =\frac{1}{2 \pi i} \int_{C\left(z_{0}, r_{1}\right)}\left[\sum_{n=1}^{\infty} f(w) \frac{\left(w-z_{0}\right)^{n-1}}{\left(z-z_{0}\right)^{n}}\right] d w \end{aligned}\]
By the Weierstrass \(M\)-test, the series converges absolutely and uniformly for \(w \in C\left(z_{0}, r_{1}\right)\).
Consequently, we may integrate term by term to obtain the series \[-\frac{1}{2 \pi i} \int_{C\left(z_{0}, r_{1}\right)} \frac{f(w)}{w-z} d w=\sum_{n=1}^{\infty} b_{n}\left(z-z_{0}\right)^{-n},\] where \[b_{n}=\frac{1}{2 \pi i} \int_{C\left(z_{0}, r_{1}\right)} \frac{f(w)}{\left(w-z_{0}\right)^{-n+1}} d w.\]
This is a power series in \(1 /\left(z-z_{0}\right)\), that converges for \(z\in \overline{D}(z_0, r_1)^c\), and hence uniformly on sets of the form \[\left\{z\in \mathbb C:\left|z-z_{0}\right| \geq 1 / \rho\right\},\] where \((1 / \rho)>r_{1}\). It follows that the convergence is uniform on compact subsets of \(\left\{z\in \mathbb C:\left|z-z_{0}\right| >r_1\right\}\).
The existence part of the theorem now follows from the above computations, and the fact that if \(s_{1}<r<s_{2}\) and \(k\in \mathbb Z\), then \[\begin{aligned} \int_{C\left(z_{0}, r\right)} \frac{f(w)d w}{\left(w-z_{0}\right)^{k+1}} &=\int_{C\left(z_{0}, r_{1}\right)} \frac{f(w)d w}{\left(w-z_{0}\right)^{k+1}}=\int_{C\left(z_{0}, r_{2}\right)} \frac{f(w)d w}{\left(w-z_{0}\right)^{k+1}}. \end{aligned}\]
We turn now to the question of uniqueness. Let \((c_{n})_{n\in\mathbb Z}\) be a sequence such that \[f(z)= \sum_{n\in\mathbb Z} c_{n}\left(z-z_{0}\right)^{n}\] for \(z \in A\left(z_{0}, s_{1}, s_{2}\right)\).
This series must converge uniformly on compact subsets of \(A\left(z_{0}, s_{1}, s_{2}\right)\). Why?
Therefore if \(k\) is any integer and \(s_{1}<r<s_{2}\), then \[\begin{aligned} &\frac{1}{2 \pi i} \int_{C\left(z_{0}, r\right)} \frac{f(w)d w}{\left(w-z_{0}\right)^{k+1}} \\ &=\frac{1}{2 \pi i} \int_{C\left(z_{0}, r\right)}\left[\sum_{n=-\infty}^{\infty} c_{n}\left(w-z_{0}\right)^{n-k-1}\right] d w \\ & =\sum_{n=-\infty}^{\infty} c_{n} \frac{1}{2 \pi i} \int_{C\left(z_{0}, r\right)}\left(w-z_{0}\right)^{n-k-1} d w \\ & =c_{k}, \end{aligned}\] because \[\frac{1}{2 \pi i} \int_{C\left(z_{0}, r\right)}\left(w-z_{0}\right)^{n-k-1} d w= \begin{cases} 1 & \text { if } n-k-1=-1, \\ 0 & \text { otherwise. } \end{cases}\] The theorem is now proved.$$\tag*{$\blacksquare$}$$