MH1810 Math

Chapter 1: Complex Numbers

Imron Rosyadi

The Need for Complex Numbers

Does \(x^2 + 1 = 0\) have a real root?

• For any real number \(x\), \(x^2 \ge 0\).
• Therefore, \(x^2 + 1 \ge 1\).
• So, \(x^2 + 1 = 0\) has no real solution.

Important

To solve equations like \(x^2 + 1 = 0\), we introduce a new concept: complex numbers.

Introducing the Imaginary Unit i

To solve \(x^2 + 1 = 0\), we define a new symbol \(i\): \(i^2 = -1\)

• Now, \(x^2 + 1 = 0\) has two distinct roots: \(i\) and \(-i\).
  • \((i)^2 + 1 = -1 + 1 = 0\)
  • \((-i)^2 + 1 = (-1)^2 (i)^2 + 1 = 1 \cdot (-1) + 1 = 0\)

Note

The imaginary unit \(i\) extends our number system beyond the real numbers.

Emphasize that this is a definition to address a mathematical “hole”. It’s not magic, but a logical extension. Mention that \(i\) is often called “iota”.

Powers of \(i\)

The powers of \(i\) follow a cyclical pattern.

\(i^1 = i\)
\(i^2 = -1\)
\(i^3 = i^2 \cdot i = -i\)
\(i^4 = i^2 \cdot i^2 = (-1)(-1) = 1\)
\(i^5 = i^4 \cdot i = 1 \cdot i = i\)

Tip

The value of \(i^n\) depends on the remainder when \(n\) is divided by 4.

For any integer \(k\):
\(i^{4k} = 1\)
\(i^{4k+1} = i\)
\(i^{4k+2} = -1\)
\(i^{4k+3} = -i\)

Go through an example: What is \(i^{27}\)? \(27 \div 4 = 6\) remainder \(3\). So \(i^{27} = i^{4 \cdot 6 + 3} = i^3 = -i\). This pattern is very useful for simplifying expressions.

Definition of a Complex Number

A complex number \(z\) is defined as:
\(z = x + yi\)
where \(x\) and \(y\) are real numbers.

• \(x\) is the real part of \(z\), denoted \(\operatorname{Re}(z)\).
• \(y\) is the imaginary part of \(z\), denoted \(\operatorname{Im}(z)\).

The set of all complex numbers is denoted by \(\mathbb{C}\).

Examples:

• \(3 + 5i\)
  • \(\operatorname{Re}(3 + 5i) = 3\)
  • \(\operatorname{Im}(3 + 5i) = 5\)
• \(3.5 - i\)
• \(-\sqrt{3} + i\)
• \(\pi + 9i\)
• \(10 = 10 + 0i\) (A real number is a complex number with imaginary part 0)
• \(- \sqrt{7}i = 0 - \sqrt{7}i\) (A purely imaginary number has real part 0)

Highlight that \(y\) is the imaginary part, not \(yi\). This is a common point of confusion for students. Explain how real numbers are a subset of complex numbers (\(\mathbb{R} \subseteq \mathbb{C}\)).

Equality of Complex Numbers

Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal.

Given \(z = x + yi\) and \(z' = x' + y'i\):
\(z = z'\) if and only if \(x = x'\) and \(y = y'\).
(i.e., \(\operatorname{Re}(z) = \operatorname{Re}(z')\) and \(\operatorname{Im}(z) = \operatorname{Im}(z')\))

Important

Suppose \(x\) and \(y\) are real numbers such that the two complex numbers \((2x - 3) + 5i\) and \((x + 7) - (y + 1)i\) are equal. Find the values of \(x\) and \(y\).

Solution:
Equating real parts: \(2x - 3 = x + 7 \implies x = 10\)
Equating imaginary parts: \(5 = -(y + 1) \implies 5 = -y - 1 \implies y = -6\)
Thus, \(x = 10\) and \(y = -6\).

This concept is fundamental for solving equations involving complex numbers. It links complex number equality back to systems of real equations.

The Argand Diagram

We can visualize complex numbers as points or vectors in a 2D plane called the Argand diagram.

• Each complex number \(z = x + yi\) corresponds to a unique point \((x,y)\) on the Cartesian plane.
• The horizontal axis represents the real part (\(\operatorname{Re}(z)\)).
• The vertical axis represents the imaginary part (\(\operatorname{Im}(z)\)).
• Alternatively, \(z = x + yi\) can be viewed as a vector from the origin \((0,0)\) to the point \((x,y)\).

Argand Diagram with point (x,y) representing z=x+iy

Emphasize the analogy with vectors. This helps in understanding addition and subtraction geometrically. Point out how real numbers lie on the x-axis and purely imaginary numbers lie on the y-axis.

Modulus and Argument

A complex number \(z=x+yi\) can also be described by its distance from the origin (\(r\)) and the angle it makes with the positive real axis (\(\theta\)).

• Modulus (\(|z|\) or \(r\)): The distance from the origin to the point \((x,y)\). \[|z| = r = \sqrt{x^2 + y^2}\]
• Argument (\(\arg(z)\) or \(\theta\)): The angle \(\theta\) (in radians) such that: \[x = r \cos \theta \quad \text{and} \quad y = r \sin \theta\] Also, \(\tan \theta = \frac{y}{x}\) (if \(x \neq 0\)).

Argand Diagram showing modulus (r) and argument (theta)

Clarify that \(\theta\) can have multiple values (\(\theta + 2k\pi\)).
Introduce the principal argument: \(-\pi < \theta \leq \pi\). This ensures a unique value for \(\theta\). Stress that the quadrant of \((x,y)\) is crucial when using \(\arctan(y/x)\) to find \(\theta\).

Polar and Exponential Forms

These forms are very useful for multiplication, division, and powers of complex numbers.

1. Polar Form (Trigonometric Form)
\(z = r(\cos \theta + i\sin \theta)\)
or \(z = r \operatorname{cis} \theta\)
where \(r = |z|\) and \(\theta = \arg(z)\).

2. Exponential Form
\(z = re^{i\theta}\)
This form uses Euler’s formula: \(e^{i\theta} = \cos \theta + i\sin \theta\).

Important

Let \(z = 3 - 3i\).

1. Find the modulus and principal argument of \(z\), and hence find its polar representation.
   
2. Write down the exponential form of \(z\).

Solution:

1. \(x=3, y=-3\).

\(r = |z| = \sqrt{3^2 + (-3)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}\).
\(\tan \theta = \frac{-3}{3} = -1\). Since \(x>0, y<0\), \(z\) is in Q4.

Principal argument \(\theta = -\frac{\pi}{4}\).

Polar form: \(z = 3\sqrt{2} \left(\cos\left(-\frac{\pi}{4}\right) + i\sin\left(-\frac{\pi}{4}\right)\right)\).

2. Exponential form: \(z = 3\sqrt{2} e^{-i\frac{\pi}{4}}\).

Emphasize the equivalence of the forms. Each has its advantages. Mention that the exponential form is widely used in electrical engineering and physics.

Complex Conjugate

The conjugate of a complex number \(z = x + yi\) is \(\bar{z} = x - yi\).

• Notation: \(\bar{z}\) or \(z^*\).
• Geometrically, \(\bar{z}\) is the reflection of \(z\) across the real axis on the Argand diagram.
• In polar form: If \(z = r(\cos \theta + i\sin \theta) = re^{i\theta}\), then \(\bar{z} = r(\cos(-\theta) + i\sin(-\theta)) = re^{-i\theta}\).

Argand Diagram showing z and its conjugate z-bar

Note

Properties of Conjugates:

• \(\overline{(\overline{z})} = z\)
• \(z\) is real if and only if \(z = \bar{z}\).
• \(z\) is purely imaginary if and only if \(z = -\bar{z}\).
• \(|\bar{z}| = |z|\) and \(\arg(\bar{z}) = -\arg(z)\).
• \(z \cdot \bar{z} = |z|^2 = x^2 + y^2\).

The conjugate is crucial for division of complex numbers (rationalizing the denominator). Demonstrate \(z \cdot \bar{z} = |z|^2\) with an example like \(z=3+4i\).

Addition and Subtraction

Complex numbers are added or subtracted by combining their real and imaginary parts separately.

Given \(z_1 = x_1 + y_1i\) and \(z_2 = x_2 + y_2i\):
\(z_1 \pm z_2 = (x_1 \pm x_2) + (y_1 \pm y_2)i\)

Important

1. \((3 + 5i) + (3.5 - i)\)
   \(= (3 + 3.5) + (5 - 1)i\)
   \(= 6.5 + 4i\)

2. \((-\sqrt{3} + i) - (\pi + 9i)\)
   \(= (-\sqrt{3} - \pi) + (1 - 9)i\)
   \(= (-\sqrt{3} - \pi) - 8i\)
   

Tip

Properties of Addition/Subtraction:

• Commutative: \(z_1 + z_2 = z_2 + z_1\)
• Associative: \((z_1 + z_2) + z_3 = z_1 + (z_2 + z_3)\)
• Conjugate: \(\overline{z_1 \pm z_2} = \overline{z_1} \pm \overline{z_2}\)

Geometrically, addition and subtraction of complex numbers are equivalent to vector addition and subtraction on the Argand diagram. Briefly illustrate this with a simple example.

Multiplication

To multiply complex numbers, use the distributive property (FOIL) and remember that \(i^2 = -1\).

Given \(z_1 = x_1 + y_1i\) and \(z_2 = x_2 + y_2i\):

\(z_1 \cdot z_2 = (x_1 + y_1i)(x_2 + y_2i)\)
\(= x_1x_2 + x_1y_2i + y_1x_2i + y_1y_2i^2\)
\(= (x_1x_2 - y_1y_2) + (x_1y_2 + x_2y_1)i\)

Important

Evaluate \((3 + 5i) \cdot (2 - i)\)
\(= (3)(2) + (3)(-i) + (5i)(2) + (5i)(-i)\)
\(= 6 - 3i + 10i - 5i^2\)
\(= 6 + 7i - 5(-1)\)
\(= 6 + 7i + 5\)
\(= 11 + 7i\)

This formula can look intimidating but emphasize it’s just FOIL. Remind students of the importance of \(i^2=-1\).

Multiplication in Polar Form

Multiplication becomes much simpler in polar form!

Given \(z_1 = r_1(\cos \theta_1 + i\sin \theta_1)\) and \(z_2 = r_2(\cos \theta_2 + i\sin \theta_2)\):
\(z_1 \cdot z_2 = r_1 r_2 (\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2))\)

Tip

Properties of Product:

1. Modulus of product = product of moduli: \(|z_1 z_2| = |z_1| |z_2|\)
2. Argument of product = sum of arguments: \(\arg(z_1 z_2) = \arg(z_1) + \arg(z_2)\)

Geometric Interpretation:

Multiplying \(z_1\) by \(z_2\) scales \(z_1\) by a factor of \(|z_2|\) and rotates it by an angle of \(\arg(z_2)\).

Argand Diagram showing multiplication of z1 and z2, resulting in z1z2 with scaled modulus and added argument

Highlight the power of the polar form for multiplication. This directly leads into De Moivre’s Theorem. Consider using a simple example where one complex number is \(2(\cos(\pi/6) + i\sin(\pi/6))\) and the other is \(3(\cos(\pi/3) + i\sin(\pi/3))\) to illustrate.

Division

To divide complex numbers, we “rationalize the denominator” using the complex conjugate.

Given \(z_1 = x_1 + y_1i\) and \(z_2 = x_2 + y_2i\) (\(z_2 \neq 0\)): \(\frac{z_1}{z_2} = \frac{z_1}{z_2} \cdot \frac{\overline{z_2}}{\overline{z_2}} = \frac{z_1 \overline{z_2}}{|z_2|^2}\)

Important

Evaluate \(\frac{3 + 5i}{2 - i}\)
\(= \frac{3 + 5i}{2 - i} \cdot \frac{2 + i}{2 + i}\)
\(= \frac{(3)(2) + (3)(i) + (5i)(2) + (5i)(i)}{2^2 + (-1)^2}\)
\(= \frac{6 + 3i + 10i + 5i^2}{4 + 1}\)
\(= \frac{6 + 13i - 5}{5}\)
\(= \frac{1 + 13i}{5}\)
\(= \frac{1}{5} + \frac{13}{5}i\)

Connect this process to rationalizing denominators with square roots. The purpose is to get a real number in the denominator.

Division in Polar Form

Similar to multiplication, division is simplified in polar form.

Given \(z_1 = r_1(\cos \theta_1 + i\sin \theta_1)\) and \(z_2 = r_2(\cos \theta_2 + i\sin \theta_2)\):
\(\frac{z_1}{z_2} = \frac{r_1}{r_2} (\cos(\theta_1 - \theta_2) + i\sin(\theta_1 - \theta_2))\)

Tip

Properties of Quotient:

1. Modulus of quotient = quotient of moduli: \(\left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|}\)
2. Argument of quotient = difference of arguments: \(\arg\left(\frac{z_1}{z_2}\right) = \arg(z_1) - \arg(z_2)\)

Important

Let \(z = \cos \theta + i\sin \theta\). Find \(|z|\) and show that \(\frac{1}{z} = \bar{z}\).

Solution:

\(|z| = \sqrt{\cos^2\theta + \sin^2\theta} = \sqrt{1} = 1\).
\(\frac{1}{z} = \frac{1}{\cos\theta + i\sin\theta} = \frac{1}{\cos\theta + i\sin\theta} \cdot \frac{\cos\theta - i\sin\theta}{\cos\theta - i\sin\theta}\)
\(= \frac{\cos\theta - i\sin\theta}{\cos^2\theta + \sin^2\theta} = \frac{\cos\theta - i\sin\theta}{1} = \cos\theta - i\sin\theta = \bar{z}\).

The example is important as it shows that for a complex number on the unit circle (\(|z|=1\)), its reciprocal is its conjugate. This will be useful for De Moivre’s Theorem.

The Fundamental Theorem of Algebra

This powerful theorem guarantees the existence of roots for polynomial equations within the complex number system.

Important

Theorem 1.5.1 (Fundamental Theorem of Algebra):

Every polynomial equation of the form \(a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0 = 0\) where \(a_k\) are complex coefficients, \(n \ge 1\), and \(a_n \neq 0\), has exactly \(n\) roots in the complex number system, provided each multiple root of multiplicity \(m\) is counted as \(m\) roots.

Note

This means that any polynomial with real or complex coefficients can always be factored completely into linear factors over the complex numbers.
The real number system is “incomplete” for roots (e.g., \(x^2+1=0\)), but the complex number system is “complete”.

Explain that “completeness” here refers to algebraic closure. No need to delve into the proof, as stated in the text. Contrast with the real number system where some polynomials have no real roots.

Solving Quadratic Equations with \(D < 0\)

When the discriminant \(D = b^2 - 4ac\) is negative, quadratic equations have complex roots.

For \(ax^2 + bx + c = 0\):

• If \(D > 0\): Two distinct real roots. \[x = \frac{-b \pm \sqrt{D}}{2a}\]
• If \(D = 0\): One repeated real root. \[x = \frac{-b}{2a}\]
• If \(D < 0\): Two distinct complex conjugate roots. \[x = \frac{-b \pm i\sqrt{-D}}{2a}\]

Important

Solve the quadratic equation \(2x^2 - 3x + 5 = 0\).

Solution:
Here \(a=2, b=-3, c=5\).
\(D = b^2 - 4ac = (-3)^2 - 4(2)(5) = 9 - 40 = -31\).
Since \(D < 0\), we have complex roots:
\(x = \frac{-(-3) \pm i\sqrt{-(-31)}}{2(2)} = \frac{3 \pm i\sqrt{31}}{4}\)
The two roots are \(x_1 = \frac{3}{4} + \frac{\sqrt{31}}{4}i\) and \(x_2 = \frac{3}{4} - \frac{\sqrt{31}}{4}i\).

Emphasize that the complex roots always come in conjugate pairs when the coefficients are real. This is a direct application of what we just learned about complex numbers.

Polynomials with Real Coefficients

A special property arises when a polynomial has real coefficients.

Important

Theorem 1.5.3 (Conjugate Root Theorem):

If \(p(x) = a_n x^n + \dots + a_0\) is a polynomial with real coefficients \(a_k\), and \(z\) is a root of \(p(x) = 0\), then its conjugate \(\bar{z}\) is also a root of \(p(x) = 0\).

Important

Suppose \(z_0 = 1 + i\) is a complex root of \(x^3 - x^2 + 2 = 0\).

By the Conjugate Root Theorem, \(\overline{z_0} = 1 - i\) is also a complex root.
Therefore, \((x - z_0)(x - \overline{z_0})\) is a quadratic factor of \(x^3 - x^2 + 2\).
\((x - (1+i))(x - (1-i)) = ((x-1) - i)((x-1) + i)\)
\(= (x-1)^2 - i^2 = (x-1)^2 + 1 = x^2 - 2x + 1 + 1 = x^2 - 2x + 2\).
This quadratic factor \(x^2 - 2x + 2\) has real coefficients.

Important

Theorem 1.5.4: Every odd degree polynomial \(p(x)\) with real coefficients has at least one real root.

Explain why the quadratic factor has real coefficients. This is key to understanding the theorem. For the odd degree theorem, if complex roots come in pairs, an odd number of roots must mean at least one is real. Example: \(9x^5 + 7x^2 - 6x + \pi = 0\) must have at least one real root because it’s an odd-degree polynomial with real coefficients.

De Moivre’s Theorem

This theorem provides a powerful way to raise complex numbers in polar form to a power.

Important

Theorem 1.6.1 (De Moivre’s Theorem):

For every integer \(n\), \((\cos \theta + i \sin \theta)^n = \cos n\theta + i \sin n\theta\)

Important

1. Express \((\cos \theta + i \sin \theta)^9\) in the form \(\cos n\theta + i\sin n\theta\).

Solution: Using De Moivre’s Theorem, \((\cos \theta + i \sin \theta)^9 = \cos 9\theta + i\sin 9\theta\).

2. Simplify \((\cos \frac{\pi}{4} + i \sin \frac{\pi}{4})^{-2}\). Solution: Using De Moivre’s Theorem, \((\cos \frac{\pi}{4} + i \sin \frac{\pi}{4})^{-2} = \cos(-2 \cdot \frac{\pi}{4}) + i\sin(-2 \cdot \frac{\pi}{4})\)

\(= \cos(-\frac{\pi}{2}) + i\sin(-\frac{\pi}{2}) = 0 - i = -i\).

Highlight that \(n\) can be any integer (positive, negative, or zero). Show the proof outline briefly using mathematical induction for positive integers, and then for negative integers by using the reciprocal and conjugate properties. The proof is elegant and reinforces previous concepts.

Proof of De Moivre’s Theorem (Case 1: \(n \ge 0\))

We prove by Mathematical Induction for \(n = 0, 1, 2, \dots\)

1. Base Case (\(n=0\)):

   \((\cos \theta + i \sin \theta)^0 = 1\)

   \(\cos(0\theta) + i \sin(0\theta) = \cos 0 + i \sin 0 = 1 + i(0) = 1\).

   The result holds for \(n=0\).

2. Inductive Hypothesis: Assume the result holds for some non-negative integer \(k\):

   \((\cos \theta + i \sin \theta)^k = \cos k\theta + i \sin k\theta\).

3. Inductive Step: Prove the result holds for \(k+1\): \((\cos \theta + i \sin \theta)^{k+1} = (\cos \theta + i \sin \theta)^k (\cos \theta + i \sin \theta)\) \(= (\cos k\theta + i \sin k\theta)(\cos \theta + i \sin \theta)\) (by Inductive Hypothesis) \(= (\cos k\theta \cos \theta - \sin k\theta \sin \theta) + i(\sin k\theta \cos \theta + \cos k\theta \sin \theta)\) \(= \cos(k\theta + \theta) + i\sin(k\theta + \theta)\) (using trigonometric sum identities) \(= \cos((k+1)\theta) + i\sin((k+1)\theta)\). Thus, by Mathematical Induction, the theorem holds for all non-negative integers \(n\).

Take time to walk through each step, especially the use of trigonometric identities for sum of angles. Emphasize the logical flow of mathematical induction.

Proof of De Moivre’s Theorem (Case 2: \(n < 0\))

Let \(n\) be a negative integer, so \(n = -m\) for some positive integer \(m\).

\((\cos \theta + i \sin \theta)^n = (\cos \theta + i \sin \theta)^{-m}\)
\(= \frac{1}{(\cos \theta + i \sin \theta)^m}\)
\(= \frac{1}{\cos m\theta + i \sin m\theta}\) (since \(m > 0\), by Case 1)
\(= \frac{1}{\cos m\theta + i \sin m\theta} \cdot \frac{\cos m\theta - i \sin m\theta}{\cos m\theta - i \sin m\theta}\) (multiplying by conjugate)
\(= \frac{\cos m\theta - i \sin m\theta}{\cos^2(m\theta) + \sin^2(m\theta)}\)
\(= \frac{\cos m\theta - i \sin m\theta}{1}\)
\(= \cos m\theta - i \sin m\theta\)
\(= \cos(-m\theta) + i \sin(-m\theta)\) (since \(\cos(-\alpha) = \cos\alpha\) and \(\sin(-\alpha) = -\sin\alpha\))
\(= \cos n\theta + i \sin n\theta\) (since \(n=-m\)).

Thus, De Moivre’s Theorem holds for all negative integers \(n\).

Highlight the use of complex conjugate and trigonometric identities for negative angles. This completes the proof for all integers \(n\).

Finding \(n\)-th Roots of a Complex Number

De Moivre’s Theorem can be extended to find roots of complex numbers.

Important

Theorem 1.7.2 (Distinct \(n\)-th roots):
For a complex number \(z = r(\cos \alpha + i \sin \alpha)\), the \(n\) distinct \(n\)-th roots are given by:
\(z_k = \sqrt[n]{r} \left(\cos \frac{\alpha + 2k\pi}{n} + i \sin \frac{\alpha + 2k\pi}{n}\right)\),
for \(k = 0, 1, 2, \dots, n-1\).

In exponential form:
\(z_k = \sqrt[n]{r} e^{i \frac{\alpha + 2k\pi}{n}}\),
for \(k = 0, 1, 2, \dots, n-1\).

Important

Find all 5th roots of \(\sqrt{3} + i\).

Solution: First, convert \(\sqrt{3} + i\) to polar form:
\(x=\sqrt{3}, y=1\).
\(r = |\sqrt{3} + i| = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3+1} = \sqrt{4} = 2\).
\(\tan \alpha = \frac{1}{\sqrt{3}}\). Since \(x>0, y>0\), \(\alpha = \frac{\pi}{6}\).
So, \(\sqrt{3} + i = 2\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right)\).

Now apply the formula for \(n=5\):
\(z_k = \sqrt[5]{2} \left(\cos \frac{\frac{\pi}{6} + 2k\pi}{5} + i \sin \frac{\frac{\pi}{6} + 2k\pi}{5}\right)\), for \(k=0,1,2,3,4\).
\(z_k = \sqrt[5]{2} \left(\cos \frac{\pi + 12k\pi}{30} + i \sin \frac{\pi + 12k\pi}{30}\right)\).

The 5 roots are:
\(k=0: z_0 = \sqrt[5]{2} (\cos \frac{\pi}{30} + i \sin \frac{\pi}{30})\)
\(k=1: z_1 = \sqrt[5]{2} (\cos \frac{13\pi}{30} + i \sin \frac{13\pi}{30})\)
\(k=2: z_2 = \sqrt[5]{2} (\cos \frac{25\pi}{30} + i \sin \frac{25\pi}{30}) = \sqrt[5]{2} (\cos \frac{5\pi}{6} + i \sin \frac{5\pi}{6})\)
\(k=3: z_3 = \sqrt[5]{2} (\cos \frac{37\pi}{30} + i \sin \frac{37\pi}{30})\)
\(k=4: z_4 = \sqrt[5]{2} (\cos \frac{49\pi}{30} + i \sin \frac{49\pi}{30})\)

Highlight the importance of adding \(2k\pi\) to the argument before dividing by \(n\). This ensures all distinct roots are found. Geometrically, the roots are equally spaced on a circle of radius \(\sqrt[n]{r}\) in the Argand diagram. The \(n\) integers can be any \(n\) consecutive integers, but \(k=0, \dots, n-1\) is standard for distinct roots.

\(n\)-th Roots of Unity

A special case of finding \(n\)-th roots is when \(z=1\). These are called the \(n\)-th roots of unity.

Since \(1 = 1 + 0i = \cos 0 + i\sin 0 = \cos 2k\pi + i\sin 2k\pi\).

Corollary 1.7.7 (n-th roots of unity):
The \(n\) distinct \(n\)-th roots of unity are:
\(z_k = \cos \frac{2k\pi}{n} + i \sin \frac{2k\pi}{n}\), for \(k = 0, 1, 2, \dots, n-1\).
By De Moivre’s Theorem, these roots can also be expressed as \(z_k = (z_1)^k\).

Geometric Interpretation:
- All \(n\)-th roots of unity lie on the unit circle (\(r=1\)) in the Argand diagram.
- They are equally spaced around the circle, with an angle of \(\frac{2\pi}{n}\) between consecutive roots.

4th roots of unity (0, pi/2, pi, 3pi/2)

(Example for \(n=4\))

Explain that \(z_0 = \cos 0 + i\sin 0 = 1\) is always one of the roots. This concept is important in many areas of mathematics and engineering, such as digital signal processing.

Deriving Trigonometric Identities (Part I)

De Moivre’s Theorem allows us to express \(\cos n\theta\), \(\sin n\theta\), and \(\tan n\theta\) in terms of powers of \(\cos \theta\), \(\sin \theta\), and \(\tan \theta\).

Tools:
\(\cos n\theta = \operatorname{Re}(\cos n\theta + i \sin n\theta) = \operatorname{Re}((\cos \theta + i \sin \theta)^n)\)
\(\sin n\theta = \operatorname{Im}(\cos n\theta + i \sin n\theta) = \operatorname{Im}((\cos \theta + i \sin \theta)^n)\)

Method:

1. Apply De Moivre’s Theorem to \((\cos \theta + i \sin \theta)^n\).
2. Expand using the binomial theorem.
3. Equate the real or imaginary parts.
4. Substitute \(\sin^2\theta = 1 - \cos^2\theta\) or \(\cos^2\theta = 1 - \sin^2\theta\) to express in desired form.

Important

Express \(\sin 3\theta\) in terms of powers of \(\sin \theta\).

Solution:
\(\sin 3\theta = \operatorname{Im}(\cos 3\theta + i \sin 3\theta) = \operatorname{Im}((\cos \theta + i \sin \theta)^3)\)
Let \(c = \cos\theta\) and \(s = \sin\theta\).
\(\operatorname{Im}((c + is)^3) = \operatorname{Im}(c^3 + 3c^2(is) + 3c(is)^2 + (is)^3)\)
\(= \operatorname{Im}(c^3 + 3ic^2s - 3cs^2 - is^3)\)
\(= \operatorname{Im}((c^3 - 3cs^2) + i(3c^2s - s^3))\)
\(= 3c^2s - s^3\)
Now substitute \(c^2 = 1 - s^2\):
\(\sin 3\theta = 3(1 - s^2)s - s^3 = 3s - 3s^3 - s^3 = 3s - 4s^3\)
\(\sin 3\theta = 3\sin\theta - 4\sin^3\theta\).

Also mention that from the same expansion, we get \(\cos 3\theta = c^3 - 3cs^2 = \cos^3\theta - 3\cos\theta\sin^2\theta\). This can be converted to \(4\cos^3\theta - 3\cos\theta\). And \(\tan 3\theta = \frac{3t - t^3}{1 - 3t^2}\) where \(t = \tan\theta\). This method systematically generates these identities.

Deriving Trigonometric Identities (Part II)

We can also express powers of \(\cos\theta\) or \(\sin\theta\) in terms of cosines and sines of multiple angles (\(\cos k\theta, \sin k\theta\)).

Main Tool: Let \(z = \cos \theta + i\sin \theta\).
Then \(\frac{1}{z} = \cos \theta - i\sin \theta = \bar{z}\).

From this, we derive:
\(z + \frac{1}{z} = (\cos \theta + i\sin \theta) + (\cos \theta - i\sin \theta) = 2\cos \theta\)
\(z - \frac{1}{z} = (\cos \theta + i\sin \theta) - (\cos \theta - i\sin \theta) = 2i\sin \theta\)

Also, by De Moivre’s Theorem:
\(z^k = \cos k\theta + i\sin k\theta\)
\(\frac{1}{z^k} = \cos k\theta - i\sin k\theta\)

Thus:
\(z^k + \frac{1}{z^k} = 2\cos k\theta\)
\(z^k - \frac{1}{z^k} = 2i\sin k\theta\)

Important

Prove that \(\cos^3\theta = \frac{1}{4}(\cos 3\theta + 3\cos \theta)\).

Proof:
Let \(z = \cos\theta + i\sin\theta\).
\(\cos^3\theta = \left(\frac{1}{2}\left(z + \frac{1}{z}\right)\right)^3\)
\(= \frac{1}{8}\left(z^3 + 3z^2\left(\frac{1}{z}\right) + 3z\left(\frac{1}{z}\right)^2 + \left(\frac{1}{z}\right)^3\right)\)
\(= \frac{1}{8}\left(z^3 + 3z + \frac{3}{z} + \frac{1}{z^3}\right)\)
\(= \frac{1}{8}\left(\left(z^3 + \frac{1}{z^3}\right) + 3\left(z + \frac{1}{z}\right)\right)\)
Using the identities above:
\(= \frac{1}{8}(2\cos 3\theta + 3(2\cos \theta))\)
\(= \frac{1}{8}(2\cos 3\theta + 6\cos \theta)\)
\(= \frac{1}{4}(\cos 3\theta + 3\cos \theta)\).

This technique is powerful for linearizing powers of sine and cosine, which is useful in integration and Fourier analysis. Connect this back to the geometric interpretation of \(z\) and \(\bar{z}\) being reflections.

Key Takeaways

• Complex Numbers Extend Reals: \(z = x + yi\), where \(i^2 = -1\), allows solving equations like \(x^2+1=0\).
• Argand Diagram: Visualizes \(z\) as a point \((x,y)\) or a vector in a 2D plane.
• Polar/Exponential Forms: \(z = r(\cos \theta + i\sin \theta) = re^{i\theta}\) are crucial for multiplication, division, and powers.
• Complex Conjugate: \(\bar{z} = x - yi\) is essential for division and has geometric meaning as a reflection.
• Operations:
  • Addition/Subtraction: Combine real and imaginary parts.
  • Multiplication/Division: Easier in polar/exponential form (multiply/divide moduli, add/subtract arguments).

• Fundamental Theorem of Algebra: Guarantees \(n\) complex roots for any \(n\)-degree polynomial.
• Conjugate Root Theorem: If a polynomial with real coefficients has a complex root \(z\), then \(\bar{z}\) is also a root.
• De Moivre’s Theorem: \((\cos \theta + i\sin \theta)^n = \cos n\theta + i\sin n\theta\) for integer \(n\).
• N-th Roots: Used to find \(n\) distinct roots of a complex number, equally spaced around a circle in the Argand diagram.
• Trigonometric Identities: Complex numbers provide a systematic way to derive advanced trig identities.

Encourage students to summarize these points in their own words and think about how they connect. This slide is a good recap for study.

Key Equations

Equation	Description
\(i^2 = -1\)	Definition of the imaginary unit.
\(z = x + yi\)	Rectangular form of a complex number, where \(x = \operatorname{Re}(z)\) and \(y = \operatorname{Im}(z)\).
\(|z| = \sqrt{x^2 + y^2}\)	Modulus (magnitude) of a complex number.
\(x = r \cos \theta, y = r \sin \theta\)	Relates rectangular and polar coordinates.
\(z = r(\cos \theta + i\sin \theta) = re^{i\theta}\)	Polar and exponential forms of a complex number.
\(\bar{z} = x - yi = r(\cos(-\theta) + i\sin(-\theta)) = re^{-i\theta}\)	Complex conjugate of \(z\).
\(z_1 \pm z_2 = (x_1 \pm x_2) + (y_1 \pm y_2)i\)	Addition/subtraction of complex numbers.
\(z_1 z_2 = (x_1x_2 - y_1y_2) + (x_1y_2 + x_2y_1)i\)	Multiplication in rectangular form.
\(z_1 z_2 = r_1 r_2 (\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2))\)	Multiplication in polar form.
\(\frac{z_1}{z_2} = \frac{z_1 \overline{z_2}}{|z_2|^2}\)	Division in rectangular form.
\(\frac{z_1}{z_2} = \frac{r_1}{r_2} (\cos(\theta_1 - \theta_2) + i\sin(\theta_1 - \theta_2))\)	Division in polar form.
\((\cos \theta + i \sin \theta)^n = \cos n\theta + i \sin n\theta\)	De Moivre’s Theorem for any integer \(n\).
\(z_k = \sqrt[n]{r} \left(\cos \frac{\alpha + 2k\pi}{n} + i \sin \frac{\alpha + 2k\pi}{n}\right)\)	\(n\)-th distinct roots of \(z = r(\cos \alpha + i\sin \alpha)\), for \(k=0, \dots, n-1\).
\(z^k + \frac{1}{z^k} = 2\cos k\theta\), \(z^k - \frac{1}{z^k} = 2i\sin k\theta\) (where \(z = \cos\theta+i\sin\theta\))	Identities for deriving trigonometric sums/powers.

Encourage students to memorize or at least understand the derivation of these key equations. Point out the connections between different forms and operations.

Key Terms

Term	Definition
Complex Number	A number of the form \(x + yi\), where \(x\) and \(y\) are real numbers, and \(i^2 = -1\).
Imaginary Unit	The symbol \(i\), defined such that \(i^2 = -1\).
Real Part	The real number \(x\) in a complex number \(z = x + yi\), denoted \(\operatorname{Re}(z)\).
Imaginary Part	The real number \(y\) in a complex number \(z = x + yi\), denoted \(\operatorname{Im}(z)\).
Argand Diagram	A graphical representation of complex numbers as points or vectors in a 2D plane.
Modulus	The distance of a complex number \(z\) from the origin in the Argand diagram, denoted \(|z|\) or \(r\).
Argument	The angle \(\theta\) that the vector representing \(z\) makes with the positive real axis in the Argand diagram, denoted \(\arg(z)\).
Principal Argument	The unique argument \(\theta\) of a complex number \(z\) such that \(-\pi < \theta \leq \pi\).
Polar Form	Representation of a complex number \(z = r(\cos \theta + i\sin \theta)\). Also called trigonometric form.
Exponential Form	Representation of a complex number \(z = re^{i\theta}\), using Euler’s formula \(e^{i\theta} = \cos \theta + i\sin \theta\).
Complex Conjugate	For a complex number \(z = x + yi\), its conjugate is \(\bar{z} = x - yi\).
Fundamental Theorem of Algebra	States that every \(n\)-degree polynomial equation has exactly \(n\) complex roots (counting multiplicity).
Conjugate Root Theorem	For a polynomial with real coefficients, if \(z\) is a root, then its conjugate \(\bar{z}\) is also a root.
De Moivre’s Theorem	A theorem that states \((\cos \theta + i \sin \theta)^n = \cos n\theta + i \sin n\theta\) for any integer \(n\).
Roots of Unity	The \(n\)-th roots of the complex number 1. These are equally spaced points on the unit circle in the Argand diagram.

Review these terms for understanding and clarity. They form the core vocabulary of complex numbers. Encourage students to define these in their own words.