Complex Numbers and Quadratic Equations

Master complex numbers, Argand diagram, and quadratic equations for JEE Mathematics.

Complex numbers extend the real number system by introducing the imaginary unit $i = \sqrt{-1}$. This allows us to solve any polynomial equation.

Overview

graph TD
    A[Complex Numbers] --> B[Representation]
    A --> C[Operations]
    A --> D[Argand Diagram]
    A --> E[Quadratic Equations]
    B --> B1[a + ib form]
    B --> B2[Polar form]
    C --> C1[Addition/Subtraction]
    C --> C2[Multiplication/Division]
    D --> D1[Modulus]
    D --> D2[Argument]
    E --> E1[Nature of Roots]
    E --> E2[Relation with Coefficients]

Complex Numbers as Ordered Pairs

A complex number $z$ is an ordered pair $(a, b)$ of real numbers, written as:

$$z = a + ib$$

where:

  • $a = \text{Re}(z)$ is the real part
  • $b = \text{Im}(z)$ is the imaginary part
  • $i = \sqrt{-1}$ is the imaginary unit

Powers of i

PowerValue
$i^0$$1$
$i^1$$i$
$i^2$$-1$
$i^3$$-i$
$i^4$$1$
$$\boxed{i^{4n} = 1, \quad i^{4n+1} = i, \quad i^{4n+2} = -1, \quad i^{4n+3} = -i}$$
JEE Tip
To find $i^n$, divide $n$ by 4 and use the remainder. For example, $i^{2023} = i^{4(505)+3} = i^3 = -i$

Algebra of Complex Numbers

Addition and Subtraction

$$(a + ib) + (c + id) = (a + c) + i(b + d)$$ $$(a + ib) - (c + id) = (a - c) + i(b - d)$$

Multiplication

$$(a + ib)(c + id) = (ac - bd) + i(ad + bc)$$

Division

$$\frac{a + ib}{c + id} = \frac{(a + ib)(c - id)}{(c + id)(c - id)} = \frac{(ac + bd) + i(bc - ad)}{c^2 + d^2}$$

Conjugate

The conjugate of $z = a + ib$ is:

$$\bar{z} = a - ib$$

Properties:

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

Argand Diagram

Complex numbers can be represented as points in a 2D plane called the Argand plane or Complex plane.

graph LR
    subgraph "Argand Plane"
    A["z = a + ib"] --> B["Point (a, b)"]
    end
    B --> C["x-axis: Real axis"]
    B --> D["y-axis: Imaginary axis"]

Modulus

The modulus (or absolute value) of $z = a + ib$ is:

$$|z| = \sqrt{a^2 + b^2}$$

This represents the distance from the origin to the point $z$.

Properties:

  • $|z| \geq 0$
  • $|z| = 0 \Leftrightarrow z = 0$
  • $|z_1 z_2| = |z_1| |z_2|$
  • $\left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|}$
  • $|z_1 + z_2| \leq |z_1| + |z_2|$ (Triangle Inequality)

Argument

The argument of $z = a + ib$ is the angle $\theta$ made with the positive real axis:

$$\arg(z) = \theta = \tan^{-1}\left(\frac{b}{a}\right)$$

The principal argument lies in $(-\pi, \pi]$.

QuadrantRange of $\theta$
I$0 < \theta < \frac{\pi}{2}$
II$\frac{\pi}{2} < \theta < \pi$
III$-\pi < \theta < -\frac{\pi}{2}$
IV$-\frac{\pi}{2} < \theta < 0$

Polar Form

A complex number can be written in polar form as:

$$z = r(\cos\theta + i\sin\theta) = re^{i\theta}$$

where $r = |z|$ and $\theta = \arg(z)$.

Euler’s Formula

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

De Moivre’s Theorem

$$(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$$
JEE Tip
Use De Moivre’s theorem for finding powers and roots of complex numbers. It’s especially useful for finding $n$th roots.

Quadratic Equations

A quadratic equation in $x$ is:

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

Quadratic Formula

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-b \pm \sqrt{D}}{2a}$$

where $D = b^2 - 4ac$ is the discriminant.

Nature of Roots

DiscriminantNature of Roots
$D > 0$Two distinct real roots
$D = 0$Two equal real roots
$D < 0$Two complex conjugate roots

For rational roots (when $a, b, c \in \mathbb{Q}$):

  • $D > 0$ and $D$ is a perfect square → Rational roots
  • $D > 0$ but not a perfect square → Irrational roots

Relations Between Roots and Coefficients

If $\alpha$ and $\beta$ are roots of $ax^2 + bx + c = 0$:

$$\boxed{\alpha + \beta = -\frac{b}{a}}$$ $$\boxed{\alpha \cdot \beta = \frac{c}{a}}$$

Formation of Quadratic Equation

Given roots $\alpha$ and $\beta$, the quadratic equation is:

$$x^2 - (\alpha + \beta)x + \alpha\beta = 0$$

Or equivalently:

$$(x - \alpha)(x - \beta) = 0$$

Important Results

Cube Roots of Unity

The solutions of $z^3 = 1$ are:

$$1, \omega = \frac{-1 + i\sqrt{3}}{2}, \omega^2 = \frac{-1 - i\sqrt{3}}{2}$$

Properties:

  • $1 + \omega + \omega^2 = 0$
  • $\omega^3 = 1$
  • $\omega \cdot \omega^2 = 1$

nth Roots of Unity

The $n$ roots of $z^n = 1$ are:

$$z_k = e^{i\frac{2\pi k}{n}} = \cos\frac{2\pi k}{n} + i\sin\frac{2\pi k}{n}, \quad k = 0, 1, 2, ..., n-1$$

Practice Problems

  1. Express $\frac{1+i}{1-i}$ in the form $a + ib$.

  2. Find the modulus and argument of $z = -1 + i\sqrt{3}$.

  3. Find all values of $z$ such that $z^4 = -16$.

  4. If $\alpha, \beta$ are roots of $x^2 - 3x + 5 = 0$, find the equation whose roots are $\alpha^2$ and $\beta^2$.

Quick Check
Can you prove that $|\bar{z}| = |z|$ for any complex number $z$?

Further Reading