Complex Numbers

A complex number is expressed as \( z = a + ib \), where \(a\) and \(b\) are real numbers, \(a\) is the real part \(\text{Re}(z)\), and \(b\) is the imaginary part \(\text{Im}(z)\). The imaginary unit \(i\) satisfies \(i^2 = -1\).

Argand Plane Representation

Complex numbers can be represented as points \(P(a,b)\) on the Argand plane, where the x-axis represents the real part and the y-axis the imaginary part.

Modulus and Argument

The modulus \(r\) of \(z\) is the distance from the origin to \(P\):

\[ r = \sqrt{a^2 + b^2} \]

The argument \(\theta\) is the angle between the positive real axis and the line segment \(OP\).

Algebra of Complex Numbers

For \(z_1 = a + ib\) and \(z_2 = c + id\):

• Addition: \(z_1 + z_2 = (a + c) + i(b + d)\)
• Subtraction: \(z_1 - z_2 = (a - c) + i(b - d)\)
• Multiplication: \(z_1 z_2 = (ac - bd) + i(ad + bc)\)
• Division: \(\frac{z_1}{z_2} = \frac{(a + ib)(c - id)}{c^2 + d^2} = \frac{(ac + bd) + i(bc - ad)}{c^2 + d^2}\)

Powers of \(i\)

The powers of \(i\) cycle every 4 steps:

• \(i^0 = 1\)
• \(i^1 = i\)
• \(i^2 = -1\)
• \(i^3 = -i\)
• \(i^4 = 1\), and so on.

Multiplicative Inverse

The inverse of a non-zero complex number \(z = a + ib\) is:

\[ z^{-1} = \frac{1}{z} = \frac{a - ib}{a^2 + b^2} \]

Multiplying \(z\) by \(z^{-1}\) yields 1.

Worked Example

Find the product and quotient of \(z_1 = 3 + 4i\) and \(z_2 = 1 - 2i\).

Solution:

Product:

\[ z_1 z_2 = (3)(1) - (4)(-2) + i[(3)(-2) + (4)(1)] = 3 + 8 + i(-6 + 4) = 11 - 2i \]

Quotient:

\[ \frac{z_1}{z_2} = \frac{(3 + 4i)(1 + 2i)}{1^2 + (-2)^2} = \frac{3 + 6i + 4i + 8i^2}{1 + 4} = \frac{3 + 10i - 8}{5} = \frac{-5 + 10i}{5} = -1 + 2i \]

Practice Set

• Level 1 – Easy: Find the modulus and argument of \(z = 1 + i\).
• Level 2 – Moderate: Compute \( (2 + 3i)^2 \) and express in standard form.
• Level 3 – Challenging: Find the multiplicative inverse of \(z = 1 - i\) and verify by multiplication.

Answer Key

• Level 1: \(r = \sqrt{1^2 + 1^2} = \sqrt{2}\), \(\theta = \frac{\pi}{4}\)
• Level 2: \((2 + 3i)^2 = 4 + 12i + 9i^2 = 4 + 12i - 9 = -5 + 12i\)
• Level 3: \(z^{-1} = \frac{1 + i}{1^2 + (-1)^2} = \frac{1 + i}{2}\). Multiplying \(z \times z^{-1} = (1 - i)(\frac{1 + i}{2}) = \frac{1 - i^2}{2} = \frac{1 + 1}{2} = 1\)

Quick Reference

Concept	Formula
Modulus	\(r = \sqrt{a^2 + b^2}\)
Argument	\(\theta = \tan^{-1}(\frac{b}{a})\)
Multiplicative Inverse	\(z^{-1} = \frac{a - ib}{a^2 + b^2}\)
Power of \(i\)	\(i^4 = 1\) (cycle every 4)

Glossary

• Complex Number: Number of the form \(a + ib\).
• Imaginary Unit: \(i\), where \(i^2 = -1\).
• Modulus: Distance from origin to point \(P(a,b)\).
• Argument: Angle with positive real axis.
• Conjugate: \(\overline{z} = a - ib\).
• Multiplicative Inverse: Number which when multiplied by \(z\) gives 1.

Quadratic Equations

A quadratic equation in variable \(x\) is of the form:

\[ ax^2 + bx + c = 0, \quad a \neq 0 \]

where \(a, b, c\) are real numbers.

Quadratic Formula

The roots of the quadratic equation are given by:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

The discriminant \(D = b^2 - 4ac\) determines the nature of roots.

Nature of Roots

• If \(D > 0\), roots are real and unequal.
• If \(D = 0\), roots are real and equal.
• If \(D < 0\), roots are complex conjugates.

Sum and Product of Roots

If roots are \(\alpha\) and \(\beta\), then:

\[ \alpha + \beta = -\frac{b}{a} \quad \text{and} \quad \alpha \beta = \frac{c}{a} \]

Solving Pure Quadratic Equations

For \(ax^2 + c = 0\),

\[ x^2 = -\frac{c}{a} \implies x = \pm \sqrt{-\frac{c}{a}} \]

Graph of Quadratic Function

The quadratic function \(f(x) = ax^2 + bx + c\) graphs as a parabola.

• If \(a > 0\), parabola opens upwards with minimum vertex.
• If \(a < 0\), parabola opens downwards with maximum vertex.

The vertex coordinates are:

\[ x = -\frac{b}{2a}, \quad y = f\left(-\frac{b}{2a}\right) = \frac{4ac - b^2}{4a} \]

Worked Example

Solve \(2x^2 - 4x + 2 = 0\) and determine the nature of roots.

Solution:

Calculate discriminant:

\[ D = (-4)^2 - 4 \times 2 \times 2 = 16 - 16 = 0 \]

Since \(D = 0\), roots are real and equal.

Roots:

\[ x = \frac{-(-4) \pm \sqrt{0}}{2 \times 2} = \frac{4}{4} = 1 \]

Practice Set

• Level 1 – Easy: Solve \(x^2 - 5x + 6 = 0\).
• Level 2 – Moderate: Find roots of \(3x^2 + 2x + 1 = 0\) and state their nature.
• Level 3 – Challenging: For \(ax^2 + bx + c = 0\), prove \((\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta\) where \(\alpha, \beta\) are roots.

Answer Key

• Level 1: Roots are \(x=2\) and \(x=3\), real and unequal.
• Level 2: \(D = 4 - 12 = -8 < 0\), roots are complex: \(x = \frac{-2 \pm i\sqrt{8}}{6} = \frac{-1}{3} \pm \frac{i\sqrt{2}}{3}\).
• Level 3: Proof:

Given roots \(\alpha, \beta\),

\[ (\alpha - \beta)^2 = \alpha^2 - 2\alpha\beta + \beta^2 \]

Also,

\[ (\alpha + \beta)^2 = \alpha^2 + 2\alpha\beta + \beta^2 \implies \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]

Substitute into first equation:

\[ (\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta \]

Quick Reference

Concept	Formula
Quadratic Equation	\(ax^2 + bx + c = 0\)
Roots	\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
Discriminant	\(D = b^2 - 4ac\)
Sum of roots	\(\alpha + \beta = -\frac{b}{a}\)
Product of roots	\(\alpha \beta = \frac{c}{a}\)
Vertex	\(\left(-\frac{b}{2a}, \frac{4ac - b^2}{4a}\right)\)

Glossary

• Quadratic Equation: Polynomial equation of degree 2.
• Discriminant: \(D = b^2 - 4ac\), determines root nature.
• Roots: Solutions of the quadratic equation.
• Vertex: Turning point of the parabola.
• Pure Quadratic Equation: Equation of form \(ax^2 + c = 0\).