MTH2003 Complex Analysis Lecture 2

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• Pre-Lecture: MTH2003 Complex Analysis Pre-Lecture 2.
• Flashcards: MTH2003 Complex Analysis Flashcards.
• Next Lecture: MTH2003 Complex Analysis Lecture 3.

How do we define Functions of a Complex Variable?

A function of a complex variable assigns each complex number $z$ in an open subset $D\subset \mathbb{C}$ a unique complex number $w$, denoted $f(z)=w$. These functions generalise real-valued functions to the complex plane and are expressed in terms of real and imaginary parts:

$$f(z)=u(x,y)+iv(x,y),\text{where }z=x+iy.$$

Examples include polynomials, exponentials, and rational functions. Understanding these functions requires familiarity with open/closed sets, boundaries, and can then be extended to multi-valued “functions”.

What is an Open Set?

An open set in $\mathbb{C}$ is a subset $D$ where every point $z_0\in D$ has a neighbourhood entirely contained within $D$. Formally:

$$\forall z_0\in D,\exists ϵ>0\text{ such that }{N}_{ϵ}(z_0)=\{z\in \mathbb{C}:|z-z_0|<ϵ\}\subset D.$$

• Key intuition: No “edge” points are included (like an interval without endpoints).
• Example: $S=\{z\in \mathbb{C}:|z-a|<r\}$ (an open disk).

What is a Closed Set?

A closed set is the complement of an open set. Equivalently, it contains all its boundary points. For example, the closure of an open disk $S$ is:

$$\overline{S}=\{z\in \mathbb{C}:|z-a|\le r\}.$$

• Key intuition: Includes its “edge”.
• Note: A set can be neither open nor closed (e.g., $\{z:0<|z|\le 1\}$ - a mixture on different regions).

What is a Boundary?

The boundary $\partial S$ of a set $S$ consists of points where every neighbourhood contains at least one point in $S$ and one not in $S$. For a disk:

$$\partial S=\{z\in \mathbb{C}:|z-a|=r\}.$$

• Key intuition: The “edge” separating interior and exterior.

How do we classify these sets?

• Open/Closed: A set is open if it has no boundary points; closed if it contains all boundary points; can also be defined with neighbourhood existence (this is the more mathematical approach).
• Bounded: Can be enclosed in a finite disk (e.g., $|z|<5$).
• Compact: Closed and bounded (critical for many theorems in complex analysis).
• Region: An open set with some boundary points added.

Examples

1. Open Disk $S_1=\{z:|z-1|<\frac{1}{4}\}$

   • Interior: All points inside the disk.
   • Boundary: $|z-1|=\frac{1}{4}$.
   • DIAGRAM HERE (circle centred at (1,0) with radius 1/4).

2. Closed Upper Half-Plane $S_2=\{z:\text{Im}(z)\ge 1\}$

   • Closed set containing its boundary $\text{Im}(z)=1$.
   • ==DIAGRAM HERE (shaded region above horizontal line y=1)==.

3. Right Half-Plane $S_3=\{z:\text{Re}(z)>0\}$

   • Open set; boundary is the imaginary axis $\text{Re}(z)=0$.
   • DIAGRAM HERE (shaded region right of the y-axis).

4. Region Above a Parabola $S_4=\{z:y>x^2\}$

   • Open set; boundary is the parabola $y=x^2$.
   • DIAGRAM HERE (shaded area above parabola).

5. Singleton Set $S_5=\{0\}$

   • Closed set (no neighbourhood of 0 is contained in $S_5$).

How do we plot these functions?

Complex functions $f(z)=u+iv$ map 2D inputs to 2D outputs. Common methods include:

1. Parametric plots: Show $u(x,y)$ and $v(x,y)$ separately as surfaces.
2. Domain colouring: Colour the domain based on $f(z)$’s argument and modulus.
3. Vector fields: Plot $(u,v)$ as vectors at each $(x,y)$.

DIAGRAM HERE (side-by-side plots: domain z-plane and image w-plane under f(z))

DIAGRAM HERE (3D plot of u(x,y) and v(x,y) as surfaces)

What about Multi-valued “Functions”?

Inverses of complex functions (e.g., roots, logarithms) are often multi-valued, requiring domain restrictions to define single-valued branches (and hence be truly valid functions).

The Logarithm

The complex logarithm $\ln z$ is multi-valued due to the periodicity of $e^z$:

$$\ln z=\ln |z|+i(\arg z+2k\pi ),k\in \mathbb{Z}.$$

We can restrict its domain by specifying $k$, normally such that $k=0$ (the principal branch):

$$\text{Log}z=\ln |z|+i\text{Arg}z:\text{Domain is }(-\pi ,\pi ]$$

Using this is a simple as substituting, for example, the principal value of $\ln i=i\frac{\pi }{2}$, as shown:

$$\ln i=\ln 1+i\left(\frac{\pi }{2}+2k\pi \right)=i\left(\frac{\pi }{2}+2k\pi \right)$$