Notes for Introduction to Complex Analysis
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Supose that the sequence \(\{z_n\}_{n=1}^{\infty}\subset \mathbb{C}\) has two limits \(z_0\) and \(w_0\). Then for any \(\epsilon >0\) choose \(n\in \mathbb{C}\) such that for any \(n\geq n_0\), we have simultaneously \[ |z_n - z_0| < \frac{\epsilon}{2}, \mbox{ and } |z_n - w_0 | < \frac{\epsilon}{2}. \] Then \[ | z_0 - w_0 | = |z_0 - z_n + z_n - w_0 | \leq | z_n - z_0 | + | z_n - w_0| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon. \] Since \(\epsilon >0\) is arbitrary, this happens only if |z₀ - w₀| = 0. Then \(z_0 = w_0\).

Let \(\epsilon >0\). Then there exists \(n_0 \in \mathbb{N}\), such that for any \(n\geq n_0\) the following holds: \[ |x_n - x_0 | = | \mbox{Re}(z_n - z_0) | \leq |z_n - z_0 | < \epsilon. \] This shows that \(\lim_{n\to \infty} x_n = x_0\). The second part follows from a similar argument on the imaginary part of the sequence.

Let \(\epsilon = 1\). By convergence, there exists \(n_0\in \mathbb{N}\) such that whenever \(k\geq n_0\) it holds that \[ |z_k - z_0 | < 1. \] Now consider \(M = \max\{ |z_1| + 1, |z_2| + 1, \ldots, |z_{k-1}| + 1, |z_0| + 1\}\). Then it is clear that \[ |z_n| \leq M, \mbox{ for all } n \in \mathbb{N}. \]

The proof relies on the same result known for sequences of Real numbers.

Let \(\{z_n\}_{n=1}^{\infty}\) a bounded sequence where \(z_n = x_n + iy_n\). Since the sequence is bounded we note that \[ |x_n | \leq |z_n| \leq M, \quad n\in\mathbb{N}. \] Then the sequence of real numbers \(\{x_n\}_{n=1}^{\infty}\) is bounded. There exists a convergent subsequence \(\left\{x_{n_k}\right\}_{k=1}^{\infty}\) with limit \(x_0\in \mathbb{R}\). Now consider the corresponding subsequence of complex numbers \(\left\{z_{n_k}\right\}_{k=1}^{\infty}\), where \(z_{n_k} = x_{n_k} + i y_{n_k}\). Since the sequence of complex numbers is bounded, we conclude that \(\left\{y_{n_k}\right\}_{k=1}^{\infty}\) is also a bounded sequence of Real numbers. Then there exists a convergent subsequence \(\left\{y_{n_{k_j}}\right\}_{j=1}^{\infty}\) wiht limit \(y_0\). Recall that any subsequence of a convergent sequence is also convergent to the same limit, and conclude that \(\left\{x_{n_{k_j}}\right\}_{j=1}^{\infty}\) has limit \(x_0\). Then \[ \lim_{j\to \infty} z_{n_{k_j}} = \lim_{j\to \infty} \left(x_{n_{k_j}} + i y_{n_{k_j}}\right) = x_0 + iy \] is a convergent subsequence of \(\{z_n\}_{n=1}^{\infty}\).

By writing \(z_n = x_n + i y_n\) and from the inequalities \[ |x_n - x_m | \leq |z_n - z_m|, \] and \[ |y_n - y_m | \leq |z_n - z_m|, \] it follows that The sequence \(\{z_n\}_{n=1}^{\infty}\) of complex numbers is Cauchy if and only if \(\{x_n\}_{n=1}^{\infty}\) and \(\{y_n\}_{n=1}^{\infty}\) are Cauchy sequences of Real numbers if and only if the sequences of real numbers \(\{x_n\}_{n=1}^{\infty}\) and \(\{y_n\}_{n=1}^{\infty}\) are convergent if and only if \(\{z_n\}_{n=1}^{\infty}\) is a convergent sequence.

Proof of 1. It follows from the triangle inequality and the definition. Let \(\epsilon > 0\), then there exists \(n_0 \in \mathbb{N}\) such that for any \(n\geq n_0\) it simultaneously holds \(|z_n - z_0| < \frac{\epsilon}{2}\) and \(|w_n - w_0| < \frac{\epsilon}{2}\), then \[ |(z_n + w_n) - (z_0 + w_0) | = |(z_n - z_0) + (w_n - w_0) | < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon. \] That is \(\displaystyle\lim_{n\to\infty}(z_n + w_n) = z_0 + w_0\).

Proof of 2. It follows also from the definition, the triangle inequality and the fact that every convergent sequence is bounded. Since \(\{z_n\}_{n=1}^{\infty}\) is convergent, there is \(M_1>0\) such that \(|z_n| < M_1\) for all \(n\in\mathbb{N}\) and the same for \(\{w_n\}_{n=1}^{\infty}\), there is \(M_2 >0\) such that \(|w_n| < M_1\) for all \(n\in\mathbb{N}\). Take \(M = \max\{M_1,M_2\}\), and then \(M>0\) is a bound for both sequences. Let \(\epsilon > 0\), then there exists \(n_0 \in \mathbb{N}\) such that for any \(n\geq n_0\) it simultaneously holds \(|z_n - z_0| < \frac{\epsilon}{2M}\) and \(|w_n - w_0| < \frac{\epsilon}{2M}\). Then

\begin{equation*} \begin{split} |z_n w_n - z_0 w_0 | &= | z_n w_n - z_n w_0 + z_n w_0 - z_0 w_0 | \\ & \leq | z_n w_n - z_n w_0 | + |z_n w_0 - z_0 w_0 | \\ & = |z_n| | w_n - w_0| + |w_0| |z_n - z_0| \\ & \leq M | w_n - w_0| + M |z_n - z_0| \\ & < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon. \end{split} \end{equation*}

That is \(\displaystyle\lim_{n\to\infty}(z_n \, w_n) = z_0 \, w_0\).

Proof of 3. It follows from a version of the triangle inequality. Let \(\epsilon > 0\), then there exists \(n_0 \in \mathbb{N}\) such that for any \(n\geq n_0\) it holds \(|z_n - z_0| < \epsilon\). Note then that \[ |\, |z_n| - |z_0| \, | \leq |z_n - z_0| < \epsilon. \] That is \(\displaystyle\lim_{n\to\infty}|z_n| = |z_0|\).

Proof of 4. Let \(\epsilon_1 = \frac{|w_0|}{2}\), then there exists \(n_0 \in \mathbb{N}\) such that for any \(n\geq n_0\) it holds \(|w_n - w_0| < \frac{|w_0|}{2}\). First note that \[ \left| \frac{1}{w_n} - \frac{1}{w_0} \right| = |\frac{ w_0 - w_n}{w_n w_0} | = \frac{ |w_0 - w_n|}{|w_n |\, |w_0|}. \]

On the other hand by the triangle inequality \[ |w_0| - |w_n| \leq \left|\, |w_0| - |w_n|\, \right| \leq |w_0 - w_n | < \frac{|w_0|}{2}, \] this implies \[ \frac{|w_0|}{2} \leq |w_n|, \] or equivalently \[ \frac{1}{|w_n|} \leq \frac{2}{|w_0|}. \] Then we get \[ \left| \frac{1}{w_n} - \frac{1}{w_0} \right| \leq \frac{ |w_0 - w_n|}{|w_n |\, |w_0|} \leq \frac{2 |w_0 - w_n|}{ |w_0|^2}. \]

For any \(\epsilon >0\) define \(\epsilon_2 = \frac{|w_0|^2\epsilon}{2}\), then there exists \(m_0 \in \mathbb{N}\) such that for any \(n\geq m_0\) it holds \(|w_n - w_0| < \frac{|w_0|^2\epsilon}{2}\). Take \(N_0 = \max\{n_0, m_0\}\). Then we can improve our last inequality, for any \(n \geq N_0\) it holds \[ \left| \frac{1}{w_n} - \frac{1}{w_0} \right| \leq \frac{ |w_0 - w_n|}{|w_n |\, |w_0|} \leq \frac{2 |w_0 - w_n|}{ |w_0|^2} < \epsilon. \] This shows that \(\displaystyle\lim_{n\to\infty}\frac{1}{w_n} = \frac{1}{w_0}\). Finally applying (2) of this proposition we have \[ \displaystyle\lim_{n\to\infty}\frac{z_0}{w_n} = \lim_{n\to\infty} z_n\frac{1}{w_n} = \lim_{n\to\infty} z_n \lim_{n\to\infty}\frac{1}{w_n} = \frac{z_0}{w_0} \]

(1 ⇒ 2) Let \(\epsilon >0\), then there is a number \(\delta >0\) such that for every \(z\in U\) with \(|z-a| < \delta\) then \(|f(z) - f(a)| < \epsilon\). This implies that if \(z\in B_{\delta}(a) \cap U\) then \(f(B_{\delta}(a) \cap U)\subseteq B_{\epsilon}(f(a))\).

(2 ⇒ 1) Let \(\epsilon >0\), then there is a number \(\delta >0\) such that \(f(B_{\delta}(a) \cap U)\subseteq B_{\epsilon}(f(a))\). This means that for every \(z\in B_{\delta}(a) \cap U\) we have \(f(z)\in B_{\epsilon} (f(a))\). This means that for every \(z\in U\) with \(|z-a| < \delta\) it holds \(|f(z) - f(a)| < \epsilon\).

(1 ⇒ 3) Let \(f(z)\) be continuous at \(a\in U\) and \(\{z_n\}_{n=1}^{\infty}\subseteq U\) with limit \(a\). For any \(\epsilon >0\) there exists a number \(\delta >0\) such that for every \(z\in U\) with \(|z-a|<\delta\) it holds \(|f(z)-f(a)| < \epsilon\). For that \(\delta >0\) there is a \(n_0\in \mathbb{N}\) such that for any \(n\geq n_0\) we have \(|z_n - a| \leq \delta\). Then by continuity this implies that \(|f(z_n) - f(a) | < \epsilon\). This shows that \(\lim_{n\to \infty} f(z_n) = f(a)\).

(3 ⇒ 1) Let \(a\in U\) and assume that for any \(\{z_n\}_{n=1}^{\infty}\subseteq U\) with limit \(a\) we have \(\lim_{n\to \infty} f(z_n) = f(a)\). Suppose on the contrary, that \(f\) is not continuous at \(a\). Then there exists a positive \(\epsilon >0\) such that for every \(\delta >0\) there is a point \(z\in U\) with \(|z -a| < \delta\) but \(|f(z) - f(a)| \geq \epsilon\). Note that this implies that for any \(n\in\mathbb{N}\) we can choose \(\delta = \frac{1}{n}\) and \(z_n\in U\) such that \(|z_n - a| < \frac{1}{n}\), but \(|f(z_n) - f(a)| \geq \epsilon\). In this way we have constructed a sequence \(\{z_n\}_{n=1}^{\infty}\) such that \(\lim_{n\to\infty}z_n = a\) but \(\lim_{n\to\infty}f(z_n) \neq f(a)\) which is a contradiction to our initial assumptions.

(1 ⇒ 3) Let \(f(z)\) be continuous at \(a\). Consider any sequence \(\{z_n\}_{n=1}^{\infty}\subseteq U\) such that \(\lim_{n\to \infty} z_n = a\). Then for \(\epsilon >0\) there is a number \(\delta >0\) such that for all \(z\in U\) with \(|z-a|<\delta\) we have \(|f(z) - f(a)| < \epsilon\). Also for such \(\delta >0\), there exists \(n_0\in \mathbb{N}\) such that for all \(n\geq n_0\) we have \(|z_n - a| < \delta\), which then implies \(|f(z_n) - f(a)| < \epsilon\).

Consider any sequence \(\{z_n\}_{n=1}^{\infty}\subseteq U_1\) such that \(\lim_{n\to \infty}z_n = z_0\). Then define \(w_n = f(z_n) \in U_2\). Since \(f\) is continuous at \(z_0\), then \(\lim_{n\to\infty}f(z_n) = \lim_{n\to\infty}w_n = w_0 = f(z_0)\). Since \(g\) is continuous at \(w_0\) then \(\lim_{n\to \infty}g(w_n) = g(w_0)\). Note that on the left hand side \(g(w_n) = g(f(z_n)) = (g\circ f)(z_n)\), and on the right hand side \(g(w_0) = g(f(z_0)) = (g\circ f)(z_0)\). This shows that \(\lim_{n\to\infty}(g\circ f)(z_n) = (g\circ f)(z_0)\).

\(\mathbb{C}\) is open. For any \(z\in \mathbb{C}\), take \(r=1\) and evidently \(B_{1}(z)\subset \mathbb{C}\).

\(\emptyset\) is open. If this wasn't true, there would exists \(z\in \emptyset\) such that for any \(r>0\), \(B_{r}(z)\nsubseteq \emptyset\). Then the statement is vacuously true.

\(B_{r}(z)\) is open. Let \(w\in B_{r}(z)\). Take \(\delta = r - |w-z|\), and note that \(\delta >0\). To show that \(B_{\delta}(w) \subset B_{r}(z)\) we need to show that any \(\xi \in B_{\delta}(w)\) is also in \(B_{r}(z)\). If \(\xi \in B_{\delta}(w)\) then \(|\xi - w| < \delta\). Since \(|\xi - z| = |\xi - w + w - z| \leq |\xi - w| + | w - z| < \delta + |w-z| = r - |w-z| + |w-z| = r\). Then \(\xi \in B_{r}(z)\).

\(\cap_{k=1}^n U_k\) is open if each \(U_i\) is open. Let \(z\in\cap_{k=1}^n U_k\). Then for each \(1\leq k \leq n\) there exists \(r_{k}>0\) such that \(B_{r_k}(z) \subseteq U_k\). Take \(r = \min\{r_1,\ldots,r_k\}\), and hence \(B_{r}(z)\subset U_k\) for every \(1\leq k \leq n\). Hence \(B_{r}(z)\subset \cap_{k=1}^n U_k\).

\(\cup_{\alpha \in I} U_{\alpha}\) is open if each \(U_{\alpha}\) is open. Let \(z\in\cup_{\alpha \in I} U_{\alpha}\). Then there exists \(\beta \in I\) such that \(z \in U_{\beta}\). Since \(U_{\beta}\) is open, there is \(r >0\) such that \(B_{r}(z)\subset U_{\beta} \subset \cup_{\alpha \in I} U_{\alpha}\).

By the previous proposition, since \(z_0\in \mbox{Cl}(A)\), we have that for every \(n\in \mathbb{N}\) we can take \(r_n = \frac{1}{n} >0\), and it holds \[ B_{r_n}(z_0)\cap A \neq \emptyset .\] Then for each \(n\) we can choose \(z_n \in B_{r_n}(z_0)\cap A\). Clearly we have \[ |z_n - z_0 | < \frac{1}{n}, \] for all \(n \in \mathbb{N}\), which implies \(\lim_{n\to\infty}z_n = z_0\).

(\(\Rightarrow\)) Let \(A\) be connected and suppose on the contrary that there is a set \(B\subset A\), that is both, open and closed, and different from \(A\) and \(\emptyset\). Then \(B\) and \(A\setminus B\) are open sets in \(A\). Then, there are open sets \(U,V\subseteq\mathbb{C}\) such that \(B = A\cap U\) and \(A\setminus B = A \cap V\). then we have:

1. \((A\cap U) \neq \emptyset\) and \((A\cap V) \neq \emptyset\).
2. \((A\cap U) \cap( A\cap V) = B \cap (A\setminus B) = \emptyset\).
3. \(A = B\cup (A\setminus B) = (A\cap U) \cup( A\cap V) = A \cap \left( U \cup V \right)\) which implies that \(A \subseteq U\cup V\).

These three points are a contradiction since \(A\) is assumed connected.

(\(\Leftarrow\)) Let \(A\) and \(\emptyset\) the only sets that are both, open and closed in \(A\). Suppose on the contrary that \(A\) is disconnected. Then there are two open sets in \(U,V\subset \mathbb{C}\) satisfying the definition above. Put \(B_1 = A\cap U\) and \(B_2 = A \cap V\). Then \(B_1\) and \(B_2\) are non-empty open sets in \(A\). Since \(B_1 \cap B_2 = \emptyset\), and \(A = B_1\cup B_2\). Then \(A\setminus B_1 = B_2\), which implies that \(B_2\) is also closed. Moreover, since \(B_1\neq \emptyset\), \(B_2 \neq A\), which is a contradiction. Then \(A\) is connected.

(\(\Rightarrow\)).

If \(K\) is compact then \(K\) is closed. Supose there is \(z_0\in \mbox{cl}(K)\) such that \(z_0\notin K\), and consider the following sets: All balls \(B_r(z_0)\) and for each \(z\in K\) choose a ball \(U_z:=B_{r_z}(z)\) such that \(r_z>0\) is small enough to not intersect one of the \(B_r(z_0)\). Note now that the collection \(\{U_z \}_{z\in K}\) is an open cover of \(K\) but clearly any finite choice of \(U_z\)'s fails to cover \(K\).

If \(K\) is compact then \(K\) is bounded. For each $z ∈ \(K\) consider the set \(U_{z} = B_{1}(z)\). Clearly \(\{U_z \}_{z\in K}\) is an open cover of \(K\), and then there is a finite open subcover \(U_{z_1}, \ldots, U_{z_n}\). Take \(M = \max\{|z_1| + 1, \ldots, |z_k| +1 \}\). Then \(K \subset \cup_{k=1}^{n} U_{z_k} \subseteq B_M\), i.e. \(K\) is bounded.

(\(\Leftarrow\)). Exercise. Hint. Show first that any closed subset of a compact set is also compact. Use the bounded property to show that \(K\) must be contained in a square of the form \(G= [-r,r]\times [-r,r]\) and show that show that \(G\) is compact.

(\(\Rightarrow\)) Let \(K\subset\mathbb{C}\) compact. If \(\{z_n\}_{n=1}^{\infty}\subset K\) is a bounded sequence since \(K\) is bounded. By Bolzano-Weierstrass' Theorem, the sequence has a convergent subsequece, say to a limit point \(z_0\in \mbox{Cl}(K)\). Since \(K\) is closed then \(z_0\in K\).

(\(\Leftarrow\)) Suppose that any \(\{z_n\}_{n=1}^{\infty}\subset K\) has a convergent subsequece. IF \(K\) is not bounded, then we can construct a sequence \(\{z_n\}_{n=1}^{\infty}\) such that \(|z_n| > n\) for all \(n\in \mathbb{N}\). But this sequence has no convergent subsequence, which is a contradiction. Then \(K\) should be bounded.

Now take any \(z_0\in \mbox{Cl}(K)\). By a previous proposition, we can construct a sequence \(\{z_{n}\}_{n=1}^{\infty}\) such that \(\lim_{n\to\infty} z_n = z_0\). By our hypothesis, the sequence \(\{z_{n_k}\}_{k=1}^{\infty}\) such that \(\lim_{k\to\infty} z_{n_k} = w_0\), for some \(w_0\in K\). By the uniqueness of the limit, \(z_0 \in K\). We he have shown \(\mbox{Cl}(K) \subseteq K\), which implies \(\mbox{Cl}(K) = K\), i.e., \(K\) is closed.

(1 \(\Rightarrow\) 2) Assume \(f\) is continuous in \(A\). Let \(V\subset \mathbb{C}\) any open such that \(f(A)\cap V \neq \emptyset\). We want to show that \(f^{-1}(V)\) is an open subset of \(A\). Let \(z_0\in f^{-1}(V)\), then \(f(z_0)\in V\), and since \(V\) is open there exists an open ball \(B_r(f(z_0)) \subset V\). Since \(f\) is assume to be continuous, if we let \(\epsilon = r\), then there exists \(\delta >0\) such that for any \(z\in A\) such that \(|z - z_0| <\delta\) we have \(|f(z) - f(z_0)| < \epsilon\). This is the same as saying that there is a \(\delta >0\) such that \(B_{\delta}(z_0)\cap A \subseteq f^{-1}\left(B_{r}(f(z_0))\right) \subset f^{-1}(V)\). Then $U_δ = which implies \(z_0\) is an interior point. Since \(z_0\) is arbitary, \(f^{-1}(V)\) is an open subset of \(A\).

(2 \(\Rightarrow\) 1) Assume now that for any \(V\subset \mathbb{C}\) we have that \(f^{-1}(V)\) is an open set in \(A\). Let \(z_0 \in A\), then for \(f(z_0) \in \mathbb{C}\), and for any \(\epsilon >0\) consider the open set \(V = B_{\epsilon}\left(f(z_0)\right)\). Then by our hypothesis, \(f^{-1}(V)\) is an open set in \(A\), which means that there is an open set \(U\subset\mathbb{C}\) such that \(U\cap A\) is completely contained in \(f^{-1}(V)\cap A\). More over, since \(z_0 \in U\) there exists \(\delta>0\) such that \(B_{\delta}(z_0)\) is contained in \(U\) and also \(f(B_{\delta}(z_0)\cap A) \subset B_{\epsilon}(f(z_0))\). The by one of our previous propositions, \(f\) is continuous at every point of \(A\).

(2 \(\Leftrightarrow\) 3) \(f^{-1}(V)\) is open in \(A\) for any \(V\in\mathbb{C}\) open \(\Leftrightarrow\) \(A \setminus f^{-1}(V) = f^{-1}(\mathbb{C}\setminus V)\) is closed in \(A\) for any closed set \(\left(\mathbb{C}\setminus V\right)\in\mathbb{C}\) \(\Leftrightarrow\) f^-1(G) is closed for any \(G\subset \mathbb{C}\) closed.

Let \(\{V_{\alpha}\}_{\alpha \in I}\) be an open cover of \(f(K)\), i.e., \(f(K) \subseteq \cup_{\alpha \in I}V_{\alpha}\). By the properties of the inverse image we have \(K \subseteq f^{-1}\left(f(K) \right) \subseteq f^{-1}\left(\cup_{\alpha \in I}V_{\alpha}\right) = \cup_{\alpha \in I}f^{-1}(V_{\alpha})\). Since \(f\) is continuous then \(f^{-1}(V_{\alpha})\) is open for every \(\alpha \in I\). This means that \(\{f^{-1}(V_{\alpha})\}_{\alpha\in I}\) is an open cover of the compact set \(K\). Then there exists a finite upen subcover \(\{f^{-1}(V_{\alpha_i})\}_{i=1}^{k}\) such that \(K\subset \cup_{i=1}^{k} f^{-1}(V_{\alpha_i})\). Now by properties of the direct image \(f(K) \subset f^{-1}\left(\cup_{i=1}^{k} f^{-1}(V_{\alpha_i}) \right) = \cup_{i=1}^{k} f\left(f^{-1}(V_{\alpha_i})\right) \subseteq \cup_{i=1}^{k} V_{\alpha_i}\). This implies that \(\{V_{\alpha_i}\}_{i=1}^{k}\) is an open subcover for \(f(K)\) and we can conclude that \(f(K)\) is compact.

The proof goes by contradiction. Let \(A\) be connected and assume that \(f(A)\) is not connected. Then, there are two open sets \(U,V \subset \mathbb{C}\) such that:

1. \((f(A)\cap U) \neq \emptyset\) and \((f(A)\cap V) \neq \emptyset\).
2. \(U\) and \(V\) are disjoint in \(\mathbb{C}\): \((f(A)\cap U) \cap( f(A)\cap V) = \emptyset\)
3. \(f(A)\subseteq U\cup V\).

By the continuity of \(f\) we have that \(f^{-1}(U)\) and \(f^{-1}(V)\) are open sets in \(A\).

1. By properties of the inverse image and using the point (1) we have \(A \cap f^{-1}(U) \neq \emptyset\) and \(A \cap f^{-1}(V) \neq \emptyset\).
   
2. Again by properties of the inverse image and point (2) we have \(\left(A\cap f^{-1}(U)\right) \cap \left( A \cap f^{-1}(V)\right) = \emptyset\).
   
3. By properties of the inverse image and point (3) we also have \(A\subseteq f^{-1}(U)\cup f^{-1}(V)\).

the last three points are in contradiction with \(A\) being connected. Hence \(f(A)\) should also be connected.

For this proof, we use the characterisation of compact sets by sequences, and by contradiction.

Suppose then that \(f\) is continuous on the compact set \(K\) but \(f\) not uniformly continuous. Then there exists \(\epsilon >0\) such that for every \(\delta >0\) there exist \(z_0,w_0\in K\) such that on one side \(|z_0 - w_0 | < \delta\) and on the other \(|f(z_0) - f(w_0)| > \epsilon\). Since this holds for any \(\delta >0\), take \(\delta_n = \frac{1}{n}\). The for each \(n\in \mathbb{N}\) there are \(z_n, w_n\in K\) such that \(|z_n - w_n | < \frac{1}{n}\) and \(|f(z_n) - f(w_n)| >0 \epsilon\).

Now consider the sequence \(\{z_n\}_{n=1}^{\infty}\subseteq K\) just constructed. By compactness we know that there exists a convergent subsequence, namely \(\{z_{n_j}\}_{j=1}^{\infty}\), such that \(\lim_{j\to \infty} z_{n_j} = a_0\).

The inequality \(| z_{n_j} - w_{n_j} | < \frac{1}{n_j}\) implies that the subsequence \(\{w_{n_j}\}_{j=1}^{\infty}\) converges also to the same limit, \(\lim_{j\to \infty} w_{n_j} = a_0\).

By the continuity of \(f\) it holds \[ \lim_{j\to \infty} f(z_{n_j}) = f(w_{n_j}) = f(a_0). \]

Then, there exits \(j_0>0\) such that for any \(j > j_0\) it holds simultaneously \[ | f(z_{n_j}) - f(a_0) | < \frac{ \epsilon}{2}, \quad \mbox{and}\quad | f(w_{n_j}) - f(a_0) | < \frac{ \epsilon}{2}. \]

The then we have the following sequence of inequalities \[ \epsilon < |f(z_{n_j}) - f(w_{n_j})| \leq | f(z_{n_j}) - f(a_0) | + | f(w_{n_j}) - f(a_0) | < \frac{ \epsilon}{2} + \frac{\epsilon}{2} = \epsilon, \] which its a contradiction itself. Consequently \(f\) is uniformly continuous in \(K\).

Note that since \(f'(z_0) = \lim_{z\to z_0}\frac{f(z)- f(z_0)}{z - z_0}\) exists and \(\lim_{z\to z_0} z - z_0 = 0\), then the limit of o the product exists and it is the product of the limits \[ \lim_{z\to z_0}f(z)- f(z_0) = \lim_{z\to z_0}\frac{f(z)- f(z_0)}{z - z_0} \cdot (z - z_0) = f'(z_0) \lim_{z\to z_0} (z - z_0) = f'(z_0) \cdot 0 = 0, \] equivalently \[ \lim_{z\to z_0}f(z) = f(z_0),\] then, \(f\) is continuous at \(z_0\).

Case 1. Suppose \(f'(z_0) \neq 0\). Then \(f(z) - f(z_0) \neq 0\) holds for a small neighbourhood around \(z_0\). We can write \[ \frac{g(f(z)) - g(f(z_0))}{z - z_0} = \frac{g(f(z)) - g(f(z_0))}{f(z) - f(z_0)} \frac{f(z)) - f(z_0)}{z - z_0}.\] Since \(f\) is complex differentiable at \(z_0\), then it is continuous and it holds \[ \lim_{z\to z_0} f(z) = f(z_0). \] Then, by taking the limit we obtain \[ \lim_{z\to z_0} \frac{g(f(z)) - g(f(z_0))}{f(z) - f(z_0)} \lim_{z\to z_0}\frac{f(z)) - f(z_0)}{z - z_0} = g'(f(z_0)) \, f(z_0). \]

Case 2. Suppose now that \(f'(z_0) = 0\). Then one can show that \(g'(f(z_0)) = 0\) and then would finish the proof. Note that since \(g'(f(z_0))\) exists, then there is a neighbourhood around \(f(z_0)\) such that the quotient \[ \frac{g(f(z)) - g(f(z_0))}{f(z) - f(z_0)} ,\] is bounded, i.e. , there is a positive number \(M>0\), and \(\delta >0\) such that for every \(0 < | f(z) - f(z_0) |< \delta\) we have \[ \left| \frac{g(f(z)) - g(f(z_0))}{f(z) - f(z_0)}\right| < M .\] Equivalently \[ |g(f(z)) - g(f(z_0))| < M | {f(z) - f(z_0)} | .\] Then \[ \left| \frac{g(f(z)) - g(f(z_0))}{z - z_0}\right| < M \left| \frac{f(z) - f(z_0)}{z - z_0} \right|.\]

This shows that \((g\circ f)'(z_0) = 0 = g'(f(z_0)) f'(z_0) = g'(f(z_0)) \cdot 0\).

(\(\Rightarrow\))

Suppose \(f\) is complex differentiable at \(z_0\). Then by definition \[ \lim_{z\to z_0} \left[\frac{f(z) - f(z_0)}{ z- z_0} - f'(z_0)\right] = 0 ,\] Then taking module we also have \[ \lim_{z\to z_0} \left|\frac{f(z) - f(z_0)}{ z- z_0} - f'(z_0)\right| = \lim_{z\to z_0} \frac{ \left|f(z) - f(z_0) - f'(z)(z-z_0)\right|}{ \left|z-z_0\right|} = 0.\]

Now writing the last expression as mapping between \(\mathbb{R}^2\), i.e, \(z= x + i y = (x, y)\), \(z_0= x_0 + i y_0 = (x_0, y_0)\), and using the matrix representation of \(f'(z_0) = a_1 + i a_2\), we can rewrite the last equaiton as follows \[ \lim_{z\to z_0} \frac{\left| f(x,y) - f(x_0,y_0) - \left( \begin{array}{cc} a_{1} & -a_{2}\\ a_{2} & a_{1}\\ \end{array} \right) \left( \begin{array}{c} x - x_0\\ y - y_0\\ \end{array} \right) \right|}{\left| (x-x_0, y-y_0) \right|} = \lim_{(x,y)\to (x_0,y_0)} \frac{\left| f(x,y) - f(x_0,y_0) - T(x - x_0, y - y_0)\right|}{\left| (x - x_0, y - y_0)\right|} = 0, \] which is clearly equivalent to \[ \lim_{h\to 0} \frac{\left| f(x_0 + h_1, y_0 + h_2) - f(x_0,y_0) - T(h)\right|}{\left| h \right|} = 0, \] hence \(f:\mathbb{R}^2 \to \mathbb{R}^2\) is real differentiable.

Then we know that if \(f(x,y) = u(x,y) + i v(x,y)\) is real differentiable at \(z_0 = (x_0,y_0)\). Then the partial derivatives of \(u\) and \(v\) exist and moreover we showed that \(Df_{z_0} = T\). Looking at the matrix representation we have \[ Df_{z_0} = \left( \begin{array}{cc} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \\ \end{array} \right) = \left( \begin{array}{cc} a_1 & -a_2 \\ a_2 & a_1 \\ \end{array} \right) = T \] which implies the Cauchy-Riemann equations: \[ \frac{\partial u}{\partial x}(x_0,y_0) = a_1 = \frac{\partial v}{\partial y}(x_0,y_0), \quad \mbox{and} \quad \frac{\partial u}{\partial y}(x_0,y_0) = - a_2 = - \frac{\partial v}{\partial x}(x_0,y_0) .\]

(\(\Leftarrow\))

Let \(u,v:\mathbb{R}^2 \to \mathbb{R}\), real differentiable functions at \(z_0 = (x_0, y_0)\), satisfying the Cauchy-Riemann equations. Define \(a_1, a_2\) by \[ \frac{\partial u}{\partial x}(x_0,y_0) = a_1 = \frac{\partial v}{\partial y}(x_0,y_0), \quad \mbox{and} \quad \frac{\partial u}{\partial y}(x_0,y_0) = - a_2 = - \frac{\partial v}{\partial x}(x_0,y_0),\] and define the following linear transformation (in matrix representation) \[ T = \left( \begin{array}{cc} a_1 & -a_2 \\ a_2 & a_1 \\ \end{array} \right). \] Now, define the complex number \(A = a_1 + i a_2\). It holds

\[ \lim_{z\to z_0} \frac{| f(z) - f(z_0) - A (z-z_0) |}{|z- z_0|} = \lim_{(x,y)\to (x_0, y_0)} \frac{| f(x,y) - f(x_0,y_0) - T ( x-x_0, y-y_0) |}{|(x,y) - (x_0, y_0)|} = 0. \]

We can conclude that \[ \lim_{z\to z_0} \frac{f(z) - f(z_0)}{z- z_0} = A, \] \(f\) is complex diffetentiable at \(z_0\).

It follows from the Cauchy-Riemann equations and the following identity \[\frac{df}{d\bar{z}} = \frac{1}{2}\left( \frac{\partial f}{\partial x} + i \frac{\partial f}{\partial y} \right) = \frac{1}{2}\left(\frac{\partial u }{\partial x} - \frac{\partial v}{\partial y}\right) + \frac{i}{2}\left(\frac{\partial u }{\partial y} + \frac{\partial v}{\partial x}\right)\]

It follows from the real case and the mean value theorem (since \(\Omega\) is connected)

Since the second-order partial derivatives are continuous, one can interchange the order of differentiation for \(u\) and \(v\). It follows then from the Cauchy-Riemann equations that \[ u_{xx} = v_{yx} = v_{xy} = -u_{yy} \Longrightarrow u_{xx} + u_{yy} = 0.\] Then, \(u\) is harmonic. The proof for \(v\) is similar.

The proof is an application of the chain rule.

Let \(\gamma_1\) and \(\gamma_2\) two curves intersecting at \(z_0\). Then the angle between the curves is the angle between the tangent vectors \(\gamma_1'(0)\) and \(\gamma_2(0)\), to compute this angle we use \[ \theta = \mbox{arg}(\gamma_1'(z)) - \mbox{arg}(\gamma_2'(z)) = \mbox{arg}\left(\frac{\gamma_1'(0)}{\gamma_2'(0)}\right). \]

The corresponding angle at \(f(z_0)\) is obtained by comparing the angles between \(f(\gamma_1 (0))' = (f\circ \gamma_1)'(0)\) and \((f\circ \gamma_2)'(0)\) We have \[ \left( f\circ \gamma_1\right)'(0) = f'(z_0) \gamma_1'(0), \] and \[ \left( f\circ \gamma_2\right)'(0) = f'(z_0) \gamma_1'(0), \] then \[ \mbox{arg}\left( \frac{\left(f\circ \gamma_1 \right)'(0)}{\left(f\circ \gamma_2\right)'(0)}\right) = \mbox{arg}\left(\frac{f'(z_0) \gamma_1'(0)}{f'(z_0) \gamma_2'(0)}\right) = \mbox{arg}\left(\frac{\gamma_1'(0)}{\gamma_2'(0)}\right) = \theta. \]

For any \(z_0\in\Omega\), if \(z_0 \neq 0\) then we will consider \(g(z) = f(z - z_0)\) so \(g(z) = f(0)\). Then its enough to prove the assertion when \(z_0 = 0\). Consider the following lines \(L_1 = \{z \in \Omega \, | \, \mbox{Im}(z) = 0, \mbox{Re}(z) \geq 0 \}\) and for some fixed \(m\in \mathbb{R}\), \(L_2 = \{ z\in \Omega \, | \, z = x + i mx, x\geq 0 \}\), where \(m=\tan \alpha\) is the angle measured from the real line and counterclockwise. Because we are assuming \(f\) is conformal at \(z=0\) the angle at \(f(0)\) between the corresponding curves \(f(L_1)\) and \(f(L_2)\) should remain unchanged.

Write \(f(z) = u(x,y) + v(x,y)\). If the angle \(\beta\) of the tangent vector at \(f(0)\) of the curve \(L_2\) is given by \(\beta\), and then angle the angle \(\beta\) of the tangent vector at \(f(0)\) of the curve \(L_1\) is given by \(\theta\), then another way to measure the angle \(\alpha\) is by noticing \(\beta - \theta = \alpha\), or equivalently \(\theta = \beta - \alpha\). Let \(c = \tan (\beta - \alpha) = \tan (\theta)\). Also notice \[ \tan(\beta)=\lim_{x\to 0} \frac{v(x,mx) - v(0,0)}{u(x,mx)- u(0,0)} = \lim_{x\to 0} \frac{\frac{v(x,mx) - v(0,0)}{x}}{\frac{u(x,mx) - u(0,0)}{x}}= \frac{v_x(0,0) + m v_y(0,0)}{u_x(0,0)+ m u_y(0,0)}, \] where we have use the assumption that \(f\) is real differentiable and the chain rule (or directional derivative). Also since we assume that \(Df_0\) is not the zero matrix, last limit is not zero and then we also know \(\cot(\beta)\).

Then

\[ c = \tan(\theta) = \tan(\beta - \alpha) = \frac{\tan(\beta) - \tan(\alpha)}{1 + \tan(\alpha)\tan(\beta)} = \frac{ \frac{v_x + m v_y}{u_x + m u_y} - m}{ 1 + m\, \frac{v_x + m v_y}{u_x + m u_y}} \]

Not then that this can be written as

\[ v_x + mv_y - m(u_x + m u_y) = c( u_x + m u_y + m v_x + m^2 v_y) \] expanding the brackets: \[ - m^2 u_y + m(v_y - u_x) + v_x = cm^2 v_y + cm( u_y + v_x) + cu_x, \]

These are polynomials in \(m\) , then the coefficients they must be identical

\begin{eqnarray} -u_y = cv_y \\ v_y - u_x = c(u_y + v_x) \\ v_x = cu_x \\ \end{eqnarray}

Then by substitution we get

\begin{equation} v_y - u_x = c(u_y + v_x) = c( - cv_y + cu_x) = -c^2 (v_y - u_x) \end{equation}

It follows \[ (1 + c^2) (v_y - u_x) = 0,\]

this implies \(u_x = v_y\). Then \(-u_y = c u_x = v_x\).

Since \(f\) is real differentiable at \(z=0\) and satisfies the Cauchy-Riemann equations then \(f\) is complex differentiable at \(z=0\).

Relying on our main assumptions:

1. If \(f\) has a continuous primitive \(F\), then by the fundamental theorem of calculus we have

\[ \int_{\alpha}f(z) dz = \int_{\alpha} F'(z) dz = \int_{a}^{b} F'(\alpha(t)) \alpha'(t) dt = \int_{a}^{b} \frac{d}{dt} F(\alpha(t)) = F(\alpha(b)) - F(\alpha(a)) = 0.\]

1. If \(f\) is \(C^1(\Omega)\), or equivalently that \(f'\) is continuous, then we apply Green's theorem and the Cauchy-Riemann's equations. We call \(R\) the open region of the complex plane bounded by the curve \(\alpha\).

Recall the Green's Theorem: \[ \int_{\alpha} (Pdx + Qdy) = \int\int_{R} (Q_x - P_y) dx\, dy.\]

Then in our case:

\begin{align*} \int_{\alpha} f(z)dz & = \int_{\alpha} (u + i v) (dx + idy) \\ &=\int_{\alpha} (u \, dx - v\, dy) + i \int_{\alpha} (v \, dx + u\, dy) \\ &= - \int\int_{R} (u_y + v_x ) dx \, dy + i \int\int_{R} (u_x - v_y) dx \, dy) \\ &= 0. \end{align*}

Let \(z_0 \in \Omega\) and define the function \[ G(z) = \left\{ \begin{array}{ccc} \frac{f(z) - f(z_0)}{z - z_0} & \mbox{for} & z \neq z_0\\ f'(z_0) & \mbox{for} & z = z_0\\ \end{array} \right. \]

Applying the Cauchy theorem for \(G\) the curve \(\alpha(t) = z_0 + e^{it}\), \(0\leq t \leq 2\pi\) it follows:

\begin{align*} 0 & = \int_{\alpha} G(z) dz \\ & = \int_{\alpha} \frac{f(z) - f(z_0)}{z-z_0} \\ &= \int_{\alpha}\frac{f(z)dz}{z-z_0} - \int_{\alpha}\frac{f(z_0) dz}{z - z_0} \\ &= \int_{\alpha}\frac{f(z)dz}{z-z_0} - f(z_0) \int_{\alpha}\frac{dz}{z - z_0}\\ &= \int_{\alpha}\frac{f(z)dz}{z-z_0} - f(z_0) \int_{0}^{2\pi}\frac{\alpha'(t)dt}{\alpha(t) - z_0}\\ &= \int_{\alpha}\frac{f(z)dz}{z-z_0} - f(z_0) \int_{0}^{2\pi}\frac{i e^{it}dt}{e^{it}}\\ &= \int_{\alpha}\frac{f(z)dz}{z-z_0} - i\, f(z_0) \int_{0}^{2\pi}dt\\ &= \int_{\alpha}\frac{f(z)dz}{z-z_0} - 2\pi\,i\, f(z_0). \end{align*}

Take any point in \(w\in B_r(z_0)\) and consider the curve \(\alpha(t) = z_0 + r e^{it}\), \(t\in [0,2\pi]\). Note then that \(|w-z_0| < r = |\alpha(t) - z_0|\). Then for any \(\xi\) such that \(|\xi - z_0| = r\) we have \[ \frac{1}{\xi - w} = \frac{1}{\xi - z_0 + z_0 - w} = \frac{1}{(\xi - z_0)\left( 1 + \frac{z_0 - w }{ \xi - z_0}\right) }= \frac{1}{\xi - z_0}\frac{1}{\left( 1 - \frac{w - z_0 }{ \xi - z_0}\right) }, \] note now that \(|w-z_0|/(\xi-z_0) < 1\), and then we can express the last term as a power series:

\[ \frac{1}{\xi - w} = \frac{1}{\xi-z_0}\sum_{k=0}^{\infty}\left( \frac{w - z_0 }{ \xi - z_0}\right)^k, \] using the fact that the series is absolutely and uniformly convergent, and the Cauchy integral formula we have: \[ f(w) = \frac{1}{2\pi i} \int_{\alpha} \frac{f(\xi)}{\xi - w}d\xi = \sum_{k = 0}^{\infty}\frac{1}{2\pi i} \int_{\alpha} \frac{f(\xi)d\xi}{(\xi - z_0)^{k+1}} (w - z_0)^k \]

Let \(z_0\in \Omega\). By the Cauchy theorem, the function \[ F(z) = \int_{z_0}^{z} f(\xi) d\xi,\] is well defined and independent of the path connecting \(z\) with \(z_0\).

By continuity of \(f\) for any \(z \in \Omega\) and \(\epsilon >0\), there exists \(\delta >0\) such that if \(|\xi - z| < \delta\) then $|f(ξ) - f(z) | < ε. Let \(|h| < \delta\), then we have \[ \frac{F(z + h) - F(z)}{h} = \frac{1}{h} \left\{ \int_{z_0}^{z + h} f(\xi)d\xi - \int_{z_0}^{z} f(\xi)d\xi \right\} = \frac{1}{h}\int_{z}^{z + h} f(\xi)d\xi. \] On the other side note that \[ f(z) = \frac{f(z)}{h} h = \frac{f(z)}{h} \int_{z}^{z+h} d\xi = \frac{1}{h} \int_{z}^{z+h} f(z) d\xi, \] then we can estimate \[ \left| \frac{F(z + h) - F(z)}{h} - f(z) \right| = \left| \frac{1}{h} \int_{z}^{z+h} (f(\xi) - f(z)) d\xi \right| \leq \frac{1}{|h|} \int_{z}^{z+h} \left|f(\xi) - f(z)\right| \left| d\xi \right| \leq \epsilon,\] where the last inequality follows when taking the straight line segment joining \(z\) with \(z + h\) (with length |h|). Then \(F'(z) = f(z)\). This implies \(f\) is analytic and hence holomorphic in \(\Omega\).

Let \(\alpha(t) = z_0 + re^{it}\) the circle around \(z_0\) and completely contained in \(\Omega\). Then by the Cauchy integral formula we have the following mean value property

\begin{align*} f(z_0) &= \frac{1}{2\pi i} \int_{\alpha} \frac{ f(z)}{z - z_0} dz \\ & = \frac{1}{2\pi i} \int_{0}^{2\pi} \frac{ f(z_0 + re^{it})}{re^{it}} ire^{it}dt \\ & = \frac{1}{2\pi } \int_{0}^{2\pi} f(z_0 + re^{it})dt. \end{align*}

We will prove that if \(z_0\in \Omega\) has maximum modulus, then \(f\) is constant, i.e., suppose \(z_0 \in \Omega\) such that \(|f(z) | \leq |f(z_0)|\) for all \(z\in \Omega\). Now consider \(A = \{ z\in \Omega \, | \, |f(z)| = |f(z_0)| \}\), note that this set is nonempty since \(z_0 \in A\).

Part I) Prove that the set \(A\) is open. Let \(w \in A\) and suppose \(B_R(w) \subset \Omega\). Then for any \(0 < r < R\) by the mean value property we have \[ |f(z_0)| = |f(w)| = \left|\frac{1}{2\pi\,i}\int_{0}^{2\pi} f(w + re^{it})dt \right| \leq \frac{1}{2\pi } \int_{0}^{2\pi} \left|f(w + re^{it})\right| dt \leq |f(z_0)| \] Then all the inequalities can be replaced by equalities and in particular we have \[ |f(w + re^{it})| = |f(z_0)|,\] for all \(0 < r < R\). this means that \(B_r(w) \subset A\), and hence, \(A\) is open.

Part II) Prove that \(A\) is closed. Recall that by topological arguments, that if \(f:\Omega \to \mathbb{C}\) is continuous, then the inverse image of a closed set in \(\mathbb{C}\) is a closed set in \(\Omega\), and similar for real valued functions. Just note that if \(r_0 = |f(z_0)|\) then \[ A = g^{-1}(\{r_0\}), \] where \(g(z) = |f(z)|\), is the composition of \(f\) with the complex norm. Since both are continuous, their composition is continuous and since \(\{r_0\}\subset\mathbb{R}\) is closed then \(A\) is closed.

Part III) Use connectedness. \(A = \Omega\). Again by topological arguments, since \(\Omega\) is assumed to be connected, the only subsets that are both open and closed are the empty set and the whole set \(\Omega\). Since \(A \neq \emptyset\) then \(A=\Omega\).

Consider the set \[ A = \{ z \in \Omega \, | \, f^{(k)}(z) = 0, k = 0,1,2,\ldots \}. \] Then show that

1. \(A \neq \emptyset\).
2. \(A\) is open.
3. \(A\) is closed.
4. Use that \(\Omega\) is connected.

Let \(z_0\in \Omega\) be a zero of \(f\) of order \(n\), then there exists a disc \(B_r(z_0)\subset \Omega\) such that for all \(z\in B_r(z_0)\) \[ f(z) = (z-z_0)^n g(z). \] Now notice that \(g(z) \neq 0\) for any \(z \in B_r(z_0)\). On the other side, note that if \(z \in B_r(z_0)\setminus \{z_0\}\), we have that \((z-z_0)\neq 0\). Then \(f(z) \neq 0\) for all \(B_r(z_0)\setminus\{z_0\}\). Hence, \(z_0\) is an isolated zero of \(f\).

Let \(F:\Omega_1 \to \mathbb{C}\) and \(G:\Omega_1\to \mathbb{C}\) be two analytic continuations to \(\Omega_1\) of \(f\).

Since \(\Omega_2 \subset \Omega_1\) is open, we can find \(z_0\in \Omega_1\) such that \(z_0\) is a limit point of \(\Omega_2\).

Then \(F\) and \(G\) satisfy the hypothesis of the Identity Principle, hence \(F=G\) on \(\Omega_1\).

Consider the function \[ G(z) = \left\{ \begin{array}{ccc} (z-w)^2 f(z) & \mbox{for} & z \in B_r(w)\setminus\{w\} \\ 0 & \mbox{for} & z = w. \\ \end{array} \right. \] Clearly \(G\) is holomorphic in \(B_r(w)\setminus\{w\}\). But also at \(w\) we have \[ \lim_{\ z \to w} \frac{ G(z) - G(w)}{z-w} = \lim_{\ z \to w} \frac{(z-w)^2 f(z)}{z-w} = \lim_{\ z \to w} (z-w)f(z) = 0, \] We claim that the last limit is equals \(0\) because, since \(f\) is bounded in \(B_r(w)\), there is \(M>0\) such that \(f(z) \leq M\) for all \(z\in B_r(w)\). Then \(0 \leq |(z-w)f(z)| \leq (z-w)M\) shows our claim.

Then \(G\) is holomorphic in \(B_r(w)\). Then since \(G(w) = 0\) and \(G'(w) = 0\) we have that its power series expansion is of the form \(G(z) = a_2 (z-w)^2 + \cdots = (z-w)^2 F(z)\). Then \(F\) is our desired function.

Take \(\epsilon >0\) small enough such that \(0 < \epsilon < \frac{R - r}{2}\), and consider the circles \(\alpha_0\) given by \(|z-z_0| = R - \epsilon\) and \(\alpha_1\) given by \(|z-z_0| = r + \epsilon\). Then for any \(w\) such that \(r + \epsilon < |w - z_0 | < R-\epsilon\), we can find a small positive number \(\delta >0\) such that \(B_\delta(w)\) is completely contained between \(\alpha_0\) and \(\alpha_1\). The function \[ g_w(\xi) = \frac{f(\xi)}{\xi - w}, \] is holomorphic in the region between \(\alpha_0\) and \(\alpha_1\) and outside \(B_\delta(w)\). Then by the observation about the Cauchy's theorem and the non-simply connected domains we can apply the Cauchy's theorem \[ 0 = \int_{\alpha_0 - \alpha_1- B_{\delta}(w)} g_w(\xi)d\xi = \int_{\alpha_0 - \alpha_1} \frac{f(\xi)}{\xi - w} d\xi - \int_{B_{\delta}(w)} g_w(\xi)d\xi = \int_{\alpha_0 - \alpha_1} \frac{f(\xi)}{\xi - w} d\xi - 2\pi i\,f(w) \]

\[ f(w) = \frac{1}{2\pi i} \int_{\alpha_0} \frac{f(\xi)}{\xi - w} d\xi - \frac{1}{2\pi i} \int_{\alpha_1} \frac{f(\xi)}{\xi - w} d\xi .\]

For the first integral, proceed as in the proof of the power extension of holomorphic functions, then \[ \frac{1}{2\pi i} \int_{\alpha_0} \frac{f(\xi)}{\xi - w} d\xi = \sum_{k=0}^{\infty} a_k (w - z_0)^k, \] with \[ a_k = \frac{1}{2\pi i } \int_{\alpha_0}\frac{f(\xi)}{(\xi - z_0)^{k+1}}, \quad k = 0,1,2,\ldots \]

For the second integral we use the geometric series, since along \(\alpha_1\) we have \(|\xi - z_0| < |w - z_0|\), then

\[ - \frac{1}{\xi - w} = - \frac{1}{\xi - z_0 + z_0 - w} = \frac{1}{(w - z_0) - ( \xi - z_0 )} = \frac{1}{w - z_0} \sum_{k=0}^{\infty}\left( \frac{\xi - z_0}{w-z_0}\right)^k,\]

Since the series is uniformly convergent then one can integrate term by term.

\[ - \frac{1}{2\pi i } \int_{\alpha_1}\frac{f(\xi)}{\xi - w}d\xi = \sum_{k=0}^{\infty} (w-z_0)^{-(k+1)} \frac{1}{2\pi i} \int_{\alpha_1} f(\xi) (\xi - z_0)^{k} d\xi = \sum_{k = -\infty}^{-1} a_k (w - z_0)^{k}, \] with the obvius definition of \(a_{k}\) for this case, namely \[ a_{k} = \frac{1}{2\pi i } \int_{\alpha_1}\frac{f(\xi)}{(\xi - z_0)^{-(k-1)}}d\xi . \]

For \(0\leq \theta \leq 2\pi\) consider the change of variable \(z = e^{i\theta}\).