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Course 1

Complex Numbers Foundations

Build the essential foundation for complex analysis by mastering complex number representations, polar form, Euler's formula, and the topology of the complex plane.

Foundation Level4-5 hours10 Practice Questions

Learning Objectives

Understand complex numbers in algebraic form z = x + iy

Master the polar representation z = r(cos θ + i sin θ) = re^(iθ)

Calculate modulus and argument of complex numbers

Apply Euler's formula and De Moivre's theorem

Understand the topology of the complex plane

Work with neighborhoods, open sets, and regions

Understand connectivity and simply connected domains

Apply complex number operations in various representations

Complex Number Representations

Definition 1.1: Complex Number

A complex number is an ordered pair

(x, y)

of real numbers, written as

z = x + iy

, where

i^2 = -1

. Here,

x = \text{Re}(z)

is called the real part and

y = \text{Im}(z)

is called the imaginary part.

The set of all complex numbers is denoted by $\mathbb{C}$ . Complex numbers can be represented in several equivalent forms, each useful for different purposes.

Definition 1.2: Modulus and Argument

For a complex number

z = x + iy \neq 0

, define:

Modulus: $|z| = \sqrt{x^2 + y^2}$
Argument: $\arg z$ is the angle $\theta$ such that $z = |z|(\cos \theta + i \sin \theta)$ , with $\arg z \in (-\pi, \pi]$ (principal argument)

Theorem 1.1: Euler's Formula

For any real number

\theta

e^{i\theta} = \cos \theta + i \sin \theta

Proof of Theorem 1.1:

This can be proven using Taylor series. The Taylor series for $e^x$ , $\cos x$ , and $\sin x$ are:

e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}

\cos x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}

\sin x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}

Substituting $x = i\theta$ into the exponential series and using $i^2 = -1$ , $i^3 = -i$ , $i^4 = 1$ , etc., we obtain:

e^{i\theta} = 1 + i\theta - \frac{\theta^2}{2!} - i\frac{\theta^3}{3!} + \frac{\theta^4}{4!} + \cdots = \cos \theta + i \sin \theta

∎

Corollary 1.1: Polar Form

Every complex number

z \neq 0

can be written as

z = re^{i\theta}

where

r = |z|

and

\theta = \arg z

. This is called the polar form of

z

Example 1.1

Problem: Express $z = 1 + i$ in polar form.

Solution:

Step 1: Compute the modulus: $|z| = \sqrt{1^2 + 1^2} = \sqrt{2}$

Step 2: Find the argument: Since $x = 1 > 0$ and $y = 1 > 0$ , we're in quadrant I. Therefore, $\arg z = \arctan(1/1) = \arctan(1) = \pi/4$

Step 3: Write in polar form: $z = \sqrt{2}e^{i\pi/4} = \sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)$

Theorem 1.2: De Moivre's Theorem

For any integer

n

and complex number

z = re^{i\theta}

z^n = r^n e^{in\theta} = r^n(\cos n\theta + i \sin n\theta)

Proof of Theorem 1.2:

Using Euler's formula and properties of exponentials: $z^n = (re^{i\theta})^n = r^n (e^{i\theta})^n = r^n e^{in\theta}$

Applying Euler's formula again: $r^n e^{in\theta} = r^n(\cos n\theta + i \sin n\theta)$

∎

Example 1.2

Problem: Compute $(1 + i)^8$ .

Solution:

From Example 1.1, $1 + i = \sqrt{2}e^{i\pi/4}$ . By De Moivre's theorem:

(1+i)^8 = (\sqrt{2})^8 e^{i8\pi/4} = 2^4 e^{i2\pi} = 16 \cdot 1 = 16

Topology of the Complex Plane

Definition 1.3: Neighborhood

An ε-neighborhood of a point

z_0 \in \mathbb{C}

is the set

N_\varepsilon(z_0) = \{z: |z - z_0| < \varepsilon\}

for some

\varepsilon > 0

Definition 1.4: Open Set

A set

D \subseteq \mathbb{C}

is open if for every

z \in D

, there exists an

\varepsilon > 0

such that

N_\varepsilon(z) \subseteq D

Definition 1.5: Connected Set

A set

D

is connected if it cannot be written as the union of two disjoint non-empty open sets. A connected open set is called a domain.

Definition 1.6: Simply Connected

A domain

D

is simply connected if every closed curve in

D

can be continuously deformed to a point without leaving

D

. Intuitively,

D

has no "holes".

Remark 1.1

Simply connected domains are crucial for Cauchy's theorem. In a simply connected domain, the integral of an analytic function along any closed curve is zero, which guarantees path independence.

Example 1.3

Problem: Determine which sets are simply connected:

$D_1 = \{z: |z| < 2\}$ (open disk)
$D_2 = \{z: 1 < |z| < 3\}$ (annulus)
$D_3 = \mathbb{C}$ (entire complex plane)

Solution:

1. $D_1$ is simply connected (no holes, can shrink any curve to a point)

2. $D_2$ is NOT simply connected (it's an annulus with a "hole" at the origin; a circle centered at the origin cannot be shrunk to a point)

3. $D_3$ is simply connected (the entire plane has no holes)

Theorem 1.3: Triangle Inequality

For any complex numbers

z_1

and

z_2

|z_1 + z_2| \leq |z_1| + |z_2|

with equality if and only if

z_1

and

z_2

have the same argument (or one is zero).

Proof of Theorem 1.3:

Consider the triangle formed by $0$ , $z_1$ , and $z_1 + z_2$ in the complex plane. The lengths of the sides are $|z_1|$ , $|z_2|$ , and $|z_1 + z_2|$ . By the geometric triangle inequality:

|z_1 + z_2| \leq |z_1| + |z_2|

Equality occurs when the vectors $z_1$ and $z_2$ point in the same direction, i.e., when they have the same argument.

∎

Corollary 1.2

For any complex numbers

z_1

and

z_2

||z_1| - |z_2|| \leq |z_1 - z_2| \leq |z_1| + |z_2|

Complex Number Operations

Theorem 1.4: Properties of Complex Conjugate

For complex numbers

z, z_1, z_2

, the conjugate

\overline{z}

satisfies:

$\overline{z_1 + z_2} = \overline{z_1} + \overline{z_2}$
$\overline{z_1 \cdot z_2} = \overline{z_1} \cdot \overline{z_2}$
$\overline{\overline{z}} = z$
$z \overline{z} = |z|^2$
$\text{Re}(z) = \frac{z + \overline{z}}{2}$ and $\text{Im}(z) = \frac{z - \overline{z}}{2i}$

Proof of Theorem 1.4:

Let $z_1 = x_1 + iy_1$ and $z_2 = x_2 + iy_2$ . Then:

\overline{z_1 + z_2} = \overline{(x_1 + x_2) + i(y_1 + y_2)} = (x_1 + x_2) - i(y_1 + y_2) = \overline{z_1} + \overline{z_2}

For multiplication: $\overline{z_1 z_2} = \overline{(x_1 x_2 - y_1 y_2) + i(x_1 y_2 + x_2 y_1)} = (x_1 x_2 - y_1 y_2) - i(x_1 y_2 + x_2 y_1)$

Expanding $\overline{z_1} \cdot \overline{z_2} = (x_1 - iy_1)(x_2 - iy_2) = x_1 x_2 - x_1 iy_2 - iy_1 x_2 + i^2 y_1 y_2 = (x_1 x_2 - y_1 y_2) - i(x_1 y_2 + x_2 y_1)$ , matching the above.

For $z \overline{z}$ : $z \overline{z} = (x + iy)(x - iy) = x^2 + y^2 = |z|^2$

∎

Example 1.4

Problem: Find all $n$ th roots of unity (solutions to $z^n = 1$ ).

Solution:

Write $1 = e^{i \cdot 0}$ in polar form. Then by De Moivre's theorem, $z = e^{i(0 + 2k\pi)/n} = e^{i2k\pi/n}$ for $k = 0, 1, 2, \ldots, n-1$ .

The $n$ th roots of unity are: $1, e^{i2\pi/n}, e^{i4\pi/n}, \ldots, e^{i2(n-1)\pi/n}$

They are equally spaced on the unit circle, forming the vertices of a regular $n$ -gon.

Advanced Topics

Definition 1.7: Riemann Sphere

The extended complex plane

\hat{\mathbb{C}} = \mathbb{C} \cup \{\infty\}

is obtained by adding a single point at infinity to

\mathbb{C}

. This can be visualized as the Riemann sphere via stereographic projection.

Remark 1.2

The Riemann sphere provides a compactification of the complex plane. Through stereographic projection, every point on the sphere corresponds to a point in the complex plane (or infinity), making the sphere a complete and compact space.

Definition 1.8: Curve in the Complex Plane

A curve in

\mathbb{C}

is a continuous function

\gamma: [a, b] \to \mathbb{C}

where

[a, b]

is a closed interval in

\mathbb{R}

. If

\gamma(a) = \gamma(b)

, the curve is closed. A curve is simple if it does not cross itself.

Definition 1.9: Jordan Curve

A Jordan curve is a simple closed curve. A Jordan curve divides the plane into two regions: an interior (bounded) and an exterior (unbounded), by the Jordan curve theorem.

Theorem 1.5: Properties of Modulus

For complex numbers

z, z_1, z_2

$|z| \geq 0$ with equality if and only if $z = 0$
$|z_1 z_2| = |z_1| |z_2|$
$\left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|}$ for $z_2 \neq 0$
$|z^n| = |z|^n$ for any integer $n$

Proof of Theorem 1.5:

The first property follows directly from the definition. For the multiplicative property, let $z_1 = r_1 e^{i\theta_1}$ and $z_2 = r_2 e^{i\theta_2}$ . Then:

z_1 z_2 = r_1 r_2 e^{i(\theta_1 + \theta_2)}

Therefore, $|z_1 z_2| = r_1 r_2 = |z_1| |z_2|$ .

For division: $\left|\frac{z_1}{z_2}\right| = \left|z_1 \cdot \frac{1}{z_2}\right| = |z_1| \cdot \left|\frac{1}{z_2}\right| = \frac{|z_1|}{|z_2|}$

∎

Example 1.5

Problem: Prove that $|z_1 - z_2|$ represents the distance between points $z_1$ and $z_2$ in the complex plane.

Solution:

Let $z_1 = x_1 + iy_1$ and $z_2 = x_2 + iy_2$ . Then:

|z_1 - z_2| = |(x_1 - x_2) + i(y_1 - y_2)| = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}

This is exactly the Euclidean distance formula between the points $(x_1, y_1)$ and $(x_2, y_2)$ in $\mathbb{R}^2$ .

Regions and Domains

Definition 1.10: Region

A region is an open, connected set in

\mathbb{C}

. A region together with some, all, or none of its boundary points is called a domain.

Remark 1.3

The distinction between "region" and "domain" varies in the literature. Some authors use "domain" to mean an open connected set. In complex analysis, both terms typically refer to open connected sets, which are essential for many theorems.

Example 1.6

Problem: Classify the following sets:

$D_1 = \{z: \text{Re}(z) > 0\}$
$D_2 = \{z: 1 < |z| < 2, \text{arg}(z) \in (0, \pi/2)\}$
$D_3 = \{z: |z| = 1\}$

Solution:

1. $D_1$ is open (the right half-plane) and connected, so it's a region. It's also simply connected (no holes).

2. $D_2$ is an open quarter-annulus in the first quadrant. It's open and connected, so it's a region. However, it's NOT simply connected because it has a "hole" (the disk $|z| \leq 1$ ).

3. $D_3$ is the unit circle. It's closed (contains its boundary) and NOT open (no interior points), so it's not a region.

Applications and Connections

Complex numbers provide powerful tools for solving real-world problems and understanding mathematical structures:

Applications of Complex Numbers

Electrical Engineering

AC circuit analysis uses complex numbers to represent impedance, with real and imaginary parts representing resistance and reactance respectively.

Signal Processing

The Fourier transform uses complex exponentials to decompose signals into frequency components, fundamental to digital signal processing.

Quantum Mechanics

Wave functions in quantum mechanics are complex-valued, and the probability amplitude involves complex conjugation.

Fluid Dynamics

Complex potential theory provides elegant solutions to two-dimensional fluid flow problems, with analytic functions describing streamlines.

Complex Number Operations - Detailed Examples

Example 1.7

Problem: Perform the division $\frac{3 + 4i}{1 - 2i}$ .

Solution:

Multiply numerator and denominator by the conjugate of the denominator:

\frac{3 + 4i}{1 - 2i} = \frac{(3 + 4i)(1 + 2i)}{(1 - 2i)(1 + 2i)} = \frac{3 + 6i + 4i + 8i^2}{1 + 4} = \frac{3 + 10i - 8}{5} = \frac{-5 + 10i}{5} = -1 + 2i

Example 1.8

Problem: Compute $(1 + i)^{10}$ .

Solution:

First, convert to polar form: $1 + i = \sqrt{2}e^{i\pi/4}$

Using De Moivre's theorem:

(1+i)^{10} = (\sqrt{2})^{10} e^{i10\pi/4} = 2^5 e^{i5\pi/2} = 32 e^{i(2\pi + \pi/2)} = 32 e^{i\pi/2} = 32i

Example 1.9

Problem: Find all cube roots of $-8$ .

Solution:

Write $-8 = 8e^{i\pi} = 8e^{i(\pi + 2k\pi)}$ for $k \in \mathbb{Z}$

The cube roots are: $w_k = 8^{1/3} e^{i(\pi + 2k\pi)/3} = 2 e^{i(\pi + 2k\pi)/3}$ for $k = 0, 1, 2$

$w_0 = 2e^{i\pi/3} = 2(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}) = 1 + i\sqrt{3}$

$w_1 = 2e^{i\pi} = -2$

$w_2 = 2e^{i5\pi/3} = 2(\cos\frac{5\pi}{3} + i\sin\frac{5\pi}{3}) = 1 - i\sqrt{3}$

Example 1.10

Problem: Compute $\sqrt{3 + 4i}$ .

Solution:

Method 1: Using polar form. First, find modulus and argument:

$|3 + 4i| = \sqrt{9 + 16} = 5$ , and since $3 > 0$ , $\arg(3 + 4i) = \arctan(4/3)$

$\sqrt{3 + 4i} = \sqrt{5} e^{i\arctan(4/3)/2}$

Method 2: Let $\sqrt{3 + 4i} = a + bi$ where $a, b \in \mathbb{R}$ . Then:

$(a + bi)^2 = a^2 - b^2 + 2abi = 3 + 4i$

Equating real and imaginary parts: $a^2 - b^2 = 3$ and $2ab = 4$ , so $ab = 2$

Solving: $a = \pm 2$ , $b = \pm 1$ . Since $ab = 2 > 0$ , both have same sign.

Solutions: $\sqrt{3 + 4i} = \pm(2 + i)$

Solving Complex Equations

Example 1.11

Problem: Solve the quadratic equation $z^2 - 2iz - 1 = 0$ .

Solution:

Using the quadratic formula: $z = \frac{2i \pm \sqrt{(-2i)^2 - 4(1)(-1)}}{2} = \frac{2i \pm \sqrt{-4 + 4}}{2} = \frac{2i \pm 0}{2} = i$

This is a double root at $z = i$ .

Example 1.12

Problem: Solve $z^2 + (1 - 2i)z - 2i = 0$ .

Solution:

Discriminant: $\Delta = (1-2i)^2 - 4(1)(-2i) = 1 - 4i - 4 + 8i = -3 + 4i$

We need $\sqrt{-3 + 4i}$ . Let $\sqrt{-3 + 4i} = a + bi$ , then:

$a^2 - b^2 = -3$ and $2ab = 4$ , so $ab = 2$

Solving gives: $a = \pm 1$ , $b = \pm 2$ . Since $ab = 2 > 0$ , $\sqrt{-3 + 4i} = \pm(1 + 2i)$

$z = \frac{-(1-2i) \pm (1 + 2i)}{2}$

Solutions: $z_1 = \frac{-1+2i + 1+2i}{2} = 2i$ and $z_2 = \frac{-1+2i - 1 - 2i}{2} = -1$

Theorem 1.6: Fundamental Theorem of Algebra (Complex Version)

Every non-constant polynomial with complex coefficients has at least one complex root. Moreover, a polynomial of degree

n

has exactly

n

roots (counting multiplicity) in the complex numbers.

Example 1.13

Problem: Find all roots of $z^3 - 1 = 0$ .

Solution:

This is the cube roots of unity. Write $1 = e^{i2k\pi}$ for $k \in \mathbb{Z}$

Roots: $z_k = e^{i2k\pi/3}$ for $k = 0, 1, 2$

$z_0 = 1$ , $z_1 = e^{i2\pi/3} = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$ , $z_2 = e^{i4\pi/3} = -\frac{1}{2} - i\frac{\sqrt{3}}{2}$

These form the vertices of an equilateral triangle on the unit circle.

Trigonometric Functions and Complex Numbers

Theorem 1.7: Euler's Formula Extensions

From Euler's formula

e^{i\theta} = \cos\theta + i\sin\theta

, we derive:

\cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2}, \quad \sin\theta = \frac{e^{i\theta} - e^{-i\theta}}{2i}

More generally, for any complex number

z

\cos z = \frac{e^{iz} + e^{-iz}}{2}, \quad \sin z = \frac{e^{iz} - e^{-iz}}{2i}

Example 1.14

Problem: Express $\cos^3\theta$ in terms of $\cos\theta$ and $\cos 3\theta$ .

Solution:

Using $\cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2}$ :

$\cos^3\theta = \left(\frac{e^{i\theta} + e^{-i\theta}}{2}\right)^3 = \frac{1}{8}(e^{i3\theta} + 3e^{i\theta} + 3e^{-i\theta} + e^{-i3\theta})$

$= \frac{1}{4}\left(\frac{e^{i3\theta} + e^{-i3\theta}}{2} + 3\cdot\frac{e^{i\theta} + e^{-i\theta}}{2}\right) = \frac{1}{4}(\cos 3\theta + 3\cos\theta)$

Therefore: $\cos^3\theta = \frac{1}{4}(\cos 3\theta + 3\cos\theta)$

Example 1.15

Problem: Find $\sin(1 + i)$ .

Solution:

Using $\sin z = \frac{e^{iz} - e^{-iz}}{2i}$ :

$\sin(1+i) = \frac{e^{i(1+i)} - e^{-i(1+i)}}{2i} = \frac{e^{i - 1} - e^{-i + 1}}{2i} = \frac{e^{-1}e^{i} - e^{1}e^{-i}}{2i}$

$= \frac{e^{-1}(\cos 1 + i\sin 1) - e(\cos 1 - i\sin 1)}{2i} = \frac{(e^{-1} - e)\cos 1 + i(e^{-1} + e)\sin 1}{2i}$

$= \frac{e + e^{-1}}{2}\sin 1 + i\frac{e - e^{-1}}{2}\cos 1 = \cosh 1 \cdot \sin 1 + i\sinh 1 \cdot \cos 1$

Corollary 1.3: De Moivre's Theorem for Trigonometric Functions

For any integer

n

\cos n\theta + i\sin n\theta = (\cos\theta + i\sin\theta)^n

Expanding the right-hand side using the binomial theorem gives formulas for

\cos n\theta

and

\sin n\theta

in terms of powers of

\cos\theta

and

\sin\theta

Geometric Applications

Example 1.16

Problem: Show that multiplication by $i$ rotates a complex number by $\pi/2$ counterclockwise.

Solution:

If $z = re^{i\theta}$ , then $iz = i \cdot re^{i\theta} = e^{i\pi/2} \cdot re^{i\theta} = re^{i(\theta + \pi/2)}$

This multiplies the argument by adding $\pi/2$ , which is a $90^\circ$ counterclockwise rotation.

Example 1.17

Problem: Find the equation of a circle in the complex plane with center $z_0$ and radius $r$ .

Solution:

The set of points at distance $r$ from $z_0$ is given by:

|z - z_0| = r

Squaring both sides: $(z - z_0)(\overline{z} - \overline{z_0}) = r^2$

If $z = x + iy$ and $z_0 = x_0 + iy_0$ , this becomes:

(x - x_0)^2 + (y - y_0)^2 = r^2

Theorem 1.8: Reflection Principle

The reflection of a point

z

across the real axis is its conjugate

\overline{z}

. More generally, reflection across a line through the origin making angle

\alpha

with the real axis is given by

e^{i2\alpha}\overline{z}

Example 1.18

Problem: Show that $z \overline{z} = |z|^2$ and use this to find $\frac{1}{z}$ in terms of $\overline{z}$ .

Solution:

If $z = x + iy$ , then $z\overline{z} = (x + iy)(x - iy) = x^2 + y^2 = |z|^2$

Therefore: $\frac{1}{z} = \frac{\overline{z}}{z\overline{z}} = \frac{\overline{z}}{|z|^2}$

This gives a formula for the multiplicative inverse using the conjugate.

Topological Properties - Extended

Definition 1.11: Boundary Point

A point

z_0

is a boundary point of a set

D

if every neighborhood of

z_0

contains both points in

D

and points not in

D

. The set of all boundary points is the boundary of

D

, denoted

\partial D

Definition 1.12: Closed Set

A set

D

is closed if it contains all its boundary points, or equivalently, if its complement is open.

Definition 1.13: Closure

The closure of a set

D

, denoted

\overline{D}

, is the union of

D

with its boundary:

\overline{D} = D \cup \partial D

Definition 1.14: Compact Set

A set

D

is compact if it is closed and bounded. In

\mathbb{C}

, this is equivalent to every sequence in

D

having a convergent subsequence with limit in

D

Example 1.19

Problem: Determine the interior, boundary, and closure of $D = \{z: 1 < |z| < 3\}$ .

Solution:

Interior: $D^\circ = D = \{z: 1 < |z| < 3\}$ (the set itself, since it's open)

Boundary: $\partial D = \{z: |z| = 1\} \cup \{z: |z| = 3\}$ (the two circles)

Closure: $\overline{D} = \{z: 1 \leq |z| \leq 3\}$ (includes the boundary)

Compact? No, because $D$ is not closed (it doesn't contain its boundary).

Theorem 1.9: Properties of Connected Sets

The union of connected sets with non-empty intersection is connected.
The continuous image of a connected set is connected.
In $\mathbb{C}$ , an open set is connected if and only if any two points can be connected by a polygonal path within the set.

Example 1.20

Problem: Is the set $D = \{z: |z| < 1\} \cup \{z: |z| > 2\}$ connected?

Solution:

No. This set consists of two disjoint open disks. There is no path connecting a point in $|z| < 1$ to a point in $|z| > 2$ that stays within $D$ .

Therefore, $D$ is not connected, but it is the union of two connected components.

Example 1.21

Problem: Compute $(2 + 3i)(4 - 5i)$ and express the result in both rectangular and polar form.

Solution:

Rectangular form: Using the distributive property:

(2 + 3i)(4 - 5i) = 2 \cdot 4 + 2 \cdot (-5i) + 3i \cdot 4 + 3i \cdot (-5i) = 8 - 10i + 12i - 15i^2 = 8 + 2i + 15 = 23 + 2i

Polar form: First compute the modulus:

|23 + 2i| = \sqrt{23^2 + 2^2} = \sqrt{529 + 4} = \sqrt{533}

The argument: $\arg(23 + 2i) = \arctan(2/23) \approx 0.0867$ radians (since both real and imaginary parts are positive).

Therefore: $23 + 2i = \sqrt{533} e^{i\arctan(2/23)}$

Example 1.22

Problem: Find all complex numbers $z$ such that $|z - 1| = |z + i|$ . What geometric object does this represent?

Solution:

Let $z = x + iy$ . Then:

|z - 1| = |(x - 1) + iy| = \sqrt{(x-1)^2 + y^2}

|z + i| = |x + i(y+1)| = \sqrt{x^2 + (y+1)^2}

Setting them equal: $\sqrt{(x-1)^2 + y^2} = \sqrt{x^2 + (y+1)^2}$

Squaring both sides: $(x-1)^2 + y^2 = x^2 + (y+1)^2$

Expanding: $x^2 - 2x + 1 + y^2 = x^2 + y^2 + 2y + 1$

Simplifying: $-2x = 2y$ , so $y = -x$

This represents a line through the origin with slope -1. Geometrically, it's the perpendicular bisector of the line segment joining $1$ and $-i$ .

Example 1.23

Problem: Show that the set $D = \{z: \text{Re}(z) > 0\}$ is a simply connected domain.

Solution:

1. Open: For any $z_0 = x_0 + iy_0$ with $x_0 > 0$ , choose $\varepsilon = x_0/2$ . Then the neighborhood $N_\varepsilon(z_0)$ is contained in $D$ because any point $z$ in this neighborhood satisfies $\text{Re}(z) > x_0 - \varepsilon = x_0/2 > 0$ .

2. Connected: Any two points in $D$ can be connected by a polygonal path (e.g., a straight line segment) that stays in $D$ since the real part remains positive along such paths.

3. Simply connected: The set $D$ is the right half-plane, which has no holes. Any closed curve in $D$ can be continuously deformed to a point without leaving $D$ .

Therefore, $D$ is a simply connected domain.

Study Tips and Common Pitfalls

1. Working with Arguments

Remember that $\arg z$ is multivalued: $\arg z = \arg_0 z + 2k\pi$ for any integer $k$ . Always specify which branch (typically $(-\pi, \pi]$ ) when computing arguments.

2. Modulus Properties

Key identity: $|z|^2 = z\overline{z}$ . Use this to simplify expressions involving moduli. Also remember: $|\text{Re}(z)| \leq |z|$ and $|\text{Im}(z)| \leq |z|$ .

3. Polar Form Calculations

For multiplication and division, polar form is often easier: $z_1 z_2 = r_1 r_2 e^{i(\theta_1 + \theta_2)}$ and $z_1/z_2 = (r_1/r_2) e^{i(\theta_1 - \theta_2)}$ . For addition, use rectangular form.

4. Common Errors

Avoid these mistakes:

Writing $\sqrt{z_1 z_2} = \sqrt{z_1}\sqrt{z_2}$ without considering branches
Assuming $\arg(z_1 z_2) = \arg z_1 + \arg z_2$ always holds (it's true up to $2\pi$ )
Forgetting that $|z_1 + z_2|$ is generally not equal to $|z_1| + |z_2|$ (only when they have the same argument)

5. Visualizing Complex Numbers

Think geometrically: addition is vector addition, multiplication scales and rotates, conjugation reflects across the real axis. Drawing diagrams helps understand many problems.

6. Topology Concepts

Key topological concepts to master:

Open sets: every point has a neighborhood contained in the set
Connected sets: cannot be split into two disjoint open subsets
Simply connected: no holes—every closed curve can shrink to a point
Boundary: the set of limit points not in the interior

7. Practice Strategy

Work through problems using both algebraic and geometric approaches. Start with simple calculations, then progress to more complex problems involving sets and topology. Practice converting between rectangular and polar forms frequently, as this skill is essential for later courses.

8. Common Topology Mistakes

Avoid these errors:

Confusing open and closed sets (a set can be neither open nor closed)
Assuming all connected sets are simply connected (annuli are connected but not simply connected)
Forgetting that the empty set and the whole space are both open and closed
Mixing up boundary and interior points

Practice Problems

Complex Numbers Foundations Practice

Questions

Correct

Accuracy

What is the modulus of the complex number

z = 3 + 4i

Easy

Not attempted

What is the principal argument of

z = -1 + i

Medium

Not attempted

Express

z = 1 + i

in polar form.

Medium

Not attempted

Using De Moivre's theorem, compute

(1 + i)^8

Hard

Not attempted

What is the conjugate of

(2 + 3i)(1 - i)

Medium

Not attempted

Find all cube roots of

-8

Hard

Not attempted

What is the triangle inequality for complex numbers?

Easy

Not attempted

A set in the complex plane is called simply connected if:

Medium

Not attempted

What is the distance between two complex numbers

z_1

and

z_2

in the complex plane?

Easy

Not attempted

Which of the following represents the set

\{z: |z - 1| < 2\}

Medium

Not attempted

Frequently Asked Questions

What is the difference between Arg z and arg z?

Arg z (with capital A) represents the set of all possible arguments of z, which is infinite: Arg z = arg z + 2kπ for k ∈ ℤ. The principal argument arg z (lowercase) is the unique value in (-π, π], representing the angle in the standard range.

Why is Euler's formula important?

Euler's formula e^(iθ) = cos θ + i sin θ connects complex exponentials with trigonometry, making complex multiplication and powers much easier to compute. It's fundamental to understanding complex functions and appears throughout complex analysis.

What does it mean for a domain to be simply connected?

A simply connected domain has 'no holes'—every closed curve can be continuously shrunk to a point without leaving the domain. This is crucial because Cauchy's theorem requires simply connected domains to guarantee path independence of integrals.

How do I find the nth roots of a complex number?

Write the number in polar form: z = re^(iθ). Then the n roots are given by r^(1/n) e^(i(θ + 2kπ)/n) for k = 0, 1, ..., n-1. These are equally spaced on a circle of radius r^(1/n).

Why do we need complex numbers?

Complex numbers complete the algebra of polynomial equations (Fundamental Theorem of Algebra), simplify many real calculations (trigonometric identities, differential equations), and provide powerful tools for analysis, physics, and engineering applications.