Complex Analysis

Lecture 12


\(*\) \(\Omega \subseteq \C\) is piecewise affinely path connected if any two points in \(\Omega\) can be connected by a finite number of line segments in \(\Omega\), joined end to end.


For open sets in \(\C\), the two definitions are equivalent

(not so in general: see for example the comb space, to be confirmed/cleaned-up).


Show: If \(\Omega_1,\Omega_2 \subseteq \C\) are open, then so is \(\Omega_1\cap \Omega_2\).

Proof: If \(\Omega_1\cap \Omega_2 = \emptyset\) done (\(\emptyset\) is open).

Otherwise, for any \(z\in \Omega_1\cap\Omega_2\) we have

\(z\in \Omega_1 \Rightarrow \exists\varepsilon_1 \gt0 \text{ such that } B_{\varepsilon_1}(z)\subset \Omega_1\) ... ♣️

\(z\in \Omega_1 \Rightarrow \exists\varepsilon_2 \gt 0 \text{ such that } B_{\varepsilon_2}(z)\subset \Omega_2\) ... 🙂


Since \(\Omega_1,\Omega_2\) are open, set \(\varepsilon = \min \{\varepsilon_1, \varepsilon_2\}\) and note that \(\varepsilon>0\).


\(B_{\varepsilon}(z)\subset \Omega_1\) by ♣️,

\(B_{\varepsilon}(z)\subset \Omega_2\) by 🙂.

So \(B_{\varepsilon}(z)\subset \Omega_1\cap\Omega_2\). Since \(z\) was arbitrary in \(\Omega_1\cap\Omega_2\).

This implies \(\text{Int } \Omega_1\cap\Omega_2= \Omega_1\cap\Omega_2\). Hence \(\Omega_1\cap\Omega_2\) is open. \(\blacksquare\)


\(*\) An open, connected subset of \(\C\) is called domain.

\(*\) An set whose interior is a domain is called a region.


\(*\) A Point \(z\in \C\) is called an accumulation point of a set \(\Omega \subseteq \C\) if every deleted neighbouhood of \(z\) intersects \(\Omega\).

Example 1: \(\Omega=\left\{\dfrac{i}{2^n}\right\}_{n\in \N}\). The only accumulation point is 0.

Example 2: \(\Omega=B_1\). The set of accumulation points is \(\conj{B_1}\).


Let \(f\) be a \(\C\)-valued function defined on a deleted neighbourhood of \(z_0\in\C\).

\(\ds \lim_{z\ra z_0}f(z)=w_0\) says: Given \(\varepsilon > 0\) exists \(\delta > 0\) such that \[0<|z-z_0|<\delta \implies \left|f(z)-w_0\right|<\varepsilon.\]

Note: \(f\) does not have to be defined at \(z_0\).



\( f(z) = \left\{ \begin{array}{ll} 0 & z\neq 0,\\ 1337 & z=0. \end{array} \right. \)    \(\ds\lim_{z\ra 0}f(z)=0\);

\[\ds \lim_{z\ra 0}\dfrac{\sin z}{z}=1.\]

Remark: If a limit exists, it is unique.

Limit Theorems

Suppose \(f(z) = f(x+iy) = u(x,y) + i v(x,y) \), \[z_0= x_0+iy_0 \text{ and } w_0 =u_0+iv_0.\]

Theorem 1: \[ \lim_{z \ra z_0} f(z)=w_0 \iff \left\{ \begin{array}{rl} \ds\lim_{(x,y)\ra (x_0, y_0)} u(x,y) & = u_0,\\ \ds\lim_{(x,y)\ra (x_0, y_0)} v(x,y) & = v_0. \end{array} \right. \]

Limit Theorems

Theorem 2: Suppose \(\ds \lim_{z \ra z_0}f(z) = w_0\) and \(\ds \lim_{z \ra z_0}g(z) = \xi_0\) and \(\lambda\in \C\). Then

  1. \(\ds \lim_{z \ra z_0}\left(f \pm g\right)(z) = w_0 \pm \xi_0\);
  2. \(\ds \lim_{z \ra z_0}\left(\lambda \cdot f \right)(z) = \lambda \cdot w_0\);
  3. \(\ds \lim_{z \ra z_0}\left(f \cdot g\right)(z) = w_0 \cdot \xi_0\);
  4. \(\ds \lim_{z \ra z_0}\dfrac{f(z)}{g(z)} = \dfrac{w_0}{\xi_0}\), as long as \(\xi_0\neq 0\).