Complete Complex Variables Notes

Functions of Complex Variables

Comprehensive Study Notes & Examples

Section 1

1. Introduction to Complex Functions

Definition: Let $f: D \subseteq \mathbb{C} \to \mathbb{C}$. If for every complex number $z \in D$, there is a corresponding complex value $w$, then $f$ is a function of a complex variable.

Notation: We write $w = f(z)$, where $z$ is the independent variable (input) and $w$ is the dependent variable (output).

The Concept of Mapping

Unlike real calculus ($y=f(x)$) which can be plotted on a single 2D plane, complex functions require two separate planes to visualize the correspondence:

$z$-plane: Represents the domain $D$ (input values).
$w$-plane: Represents the range or image of the function (output values).

Example: Let $f(z) = \frac{1}{z}$.

If we input the point $z_0 = 1+i$, we can calculate the output:

$$f(1+i) = \frac{1}{1+i} = \frac{1-i}{(1+i)(1-i)} = \frac{1-i}{2} = \frac{1}{2} – \frac{i}{2}$$

This function maps the point $(1, 1)$ in the $z$-plane to the point $(\frac{1}{2}, -\frac{1}{2})$ in the $w$-plane.

Section 2

2. Coordinate Representations

2a. Cartesian & Conjugate Coordinates

We can express complex variables and functions using standard Cartesian coordinates $(x,y)$ or Conjugate coordinates $(z, \bar{z})$. This is useful for algebraic manipulation.

Standard Form: $\quad z = x + iy \quad \text{and} \quad \bar{z} = x – iy$

Transformation Formulas: $$x = \frac{z + \bar{z}}{2}, \quad y = \frac{z – \bar{z}}{2i}$$

Application: Functional Transformation

Problem: Transform the real-variable function $f(z) = x^2 – y^2$ into complex variables $z$ and $\bar{z}$.

By substituting the transformation formulas:

$$f(z) = \left(\frac{z+\bar{z}}{2}\right)^2 – \left(\frac{z-\bar{z}}{2i}\right)^2$$

After expanding and simplifying, the result is elegant:

$$= \frac{1}{2}(z^2 + \bar{z}^2)$$

2b. Polar Coordinates

For problems involving rotation or magnitude, polar coordinates are often superior.

Polar Form: $z = re^{i\theta}$
Where $r = |z| = \sqrt{x^2+y^2}$ and $\theta = \tan^{-1}(y/x)$.

            Mapping Example: Consider $w = f(z) = (r+1)e^{i\theta}$.
            This function takes a point with distance $r$ and maps it to a point with distance $r+1$, while keeping the angle $\theta$ constant.
Domain Analysis: It maps the punctured plane $D_1 = \mathbb{C} – \{0\}$ to the exterior of the unit circle $D_2 = \{w : |w| > 1\}$.
The mapping is one-to-one and onto.

        

Section 3

3. Bounded Functions

Definition: A function $f: D \subseteq \mathbb{C} \to \mathbb{C}$ is bounded on D if there exists a real number $M$ such that $|f(z)| \leq M$ for all $z \in D$.

Testing for Boundedness

To determine if a function is bounded, we calculate its modulus $|f(z)|$ and check for behavior that tends towards infinity within the given domain.

Example 1: Exponential of a Square ($f(z) = e^{-z^2}$)

Expansion: $f(z) = e^{-(x^2 – y^2 + 2ixy)} = e^{y^2 – x^2} \cdot e^{-2ixy}$

Modulus: $|f(z)| = e^{y^2 – x^2}$

On the Real Axis ($y=0$): $|f(z)| = e^{-x^2}$. Since $e^{-x^2} \le 1$, the function is Bounded.
On the Imaginary Axis ($x=0$): $|f(z)| = e^{y^2}$. As $y \to \infty$, the value explodes. The function is Unbounded.

Example 2: Exponential on Upper Half-Plane ($f(z) = e^{iz}$)

$|e^{i(x+iy)}| = |e^{ix} \cdot e^{-y}| = e^{-y}$

If we restrict the domain to the upper half-plane ($y > 0$), then $e^{-y} < 1$. Thus, the function becomes Bounded.

Section 4

4. Component Forms ($u, v$)

A fundamental property of complex functions is that they can be split into real and imaginary parts, each being a real-valued function of two variables:

$$f(z) = u(x, y) + i v(x, y)$$

Function $f(z)$	Real Part $u(x,y)$	Imaginary Part $v(x,y)$
$z^2 = (x+iy)^2$	$x^2 – y^2$	$2xy$
$z + xy + 2\bar{z}$	$3x + xy$	$-y$

Section 5

5. The Inversion Transformation ($w = 1/z$)

The reciprocal function $f(z) = \frac{1}{z}$ is a classic example of conformal mapping (mappings that preserve angles). We analyze it using component forms.

Component Derivation

If $w = u + iv = \frac{1}{x+iy}$, we can solve for $u, v$ in terms of $x, y$ (and vice-versa):

$$u = \frac{x}{x^2+y^2}, \quad v = \frac{-y}{x^2+y^2}$$
$$x = \frac{u}{u^2+v^2}, \quad y = \frac{-v}{u^2+v^2}$$

Geometric Insight: Circle to Line

Consider the general equation of a circle or line in the $z$-plane: $a(x^2+y^2) + bx + cy + d = 0$. Substituting the components yields the image in the $w$-plane:

$$d(u^2+v^2) + bu – cv + a = 0$$

            Key Result: If $a=0$ and $d \neq 0$ (which represents a line in the $z$-plane not passing through the origin), the image equation becomes a circle passing through the origin in the $w$-plane.
        

Section 6

6. Sequences of Complex Numbers

Definition: A sequence $\langle z_n \rangle$ is defined by $z_n = a_n + i b_n$, where $\langle a_n \rangle$ and $\langle b_n \rangle$ are standard sequences of real numbers.

Convergence & Boundedness

The properties of complex sequences are directly inherited from their real components:

Boundedness: $\langle z_n \rangle$ is bounded iff both $\langle a_n \rangle$ and $\langle b_n \rangle$ are bounded.
Convergence: $\langle z_n \rangle$ converges iff both $\langle a_n \rangle$ and $\langle b_n \rangle$ converge.

Limit Rule: If $\lim a_n = a$ and $\lim b_n = b$, then:
            
                $$\lim_{n \to \infty} z_n = a + ib$$

Example Problem

Find the limit of $z_n = \cos(1/n) + i(1 – \sin(1/n))$.

Real part limit: $\lim \cos(1/n) = \cos(0) = 1$
Imaginary part limit: $\lim (1 – \sin(1/n)) = 1 – 0 = 1$

Conclusion: $\lim z_n = 1 + i$.

function of complex variable