🕉️ VEDIC MATHEMATICS — LEVEL 3: ADVANCED

MODULE 26: Complex Numbers — Vedic Approach

Complete Study Material | Theory + Examples + Practice + Test Bank

"Complex numbers are not 'imaginary'—they are a higher dimension of reality. The sutras reveal the elegant patterns hidden within their algebra and geometry." — Vedic Mathematics Teacher's Manual

📋 MODULE AT A GLANCE

🎯 LEARNING OUTCOMES

By the end of this module, the student will be able to:

1. Multiply complex numbers in under 3 seconds using Urdhva cross-multiplication
2. Divide complex numbers using Paravartya (transpose and apply)
3. Compute modulus and argument using Vedic pattern recognition
4. Apply De Moivre's theorem for powers and roots using Yavadunam patterns
5. Find nth roots of unity and their sums using cyclic pattern analysis
6. Solve complex equations using Shunyam Samya (if same, zero)
7. Interpret complex numbers geometrically on the Argand plane
8. Complete JEE-level complex number problems in 50% less time

PART 1: THEORY

1.1 — Introduction to Complex Numbers

What is a Complex Number?

A complex number is of the form $z = a + ib$, where $a$ is the real part, $b$ is the imaginary part, and $i = \sqrt{-1}$.

$$z = a + ib, \quad a,b \in \mathbb{R}, \quad i^2 = -1$$

Why Vedic Approach to Complex Numbers?

1.2 — Multiplication of Complex Numbers: Urdhva-Tiryagbhyam

The Pattern

For $z_1 = a + ib$ and $z_2 = c + id$:

$$z_1 \times z_2 = (a + ib)(c + id) = (ac - bd) + i(ad + bc)$$

This is exactly the Urdhva-Tiryagbhyam pattern:

• Real part = $ac - bd$ (product of first terms minus product of last terms)
• Imaginary part = $ad + bc$ (cross terms sum)

Visual Pattern

    (a + ib) × (c + id)
         │        │
         └──┬──┬──┘
            │  │
    ac + i²bd + i(ad + bc)
    = (ac - bd) + i(ad + bc)

Example 1: $(2 + 3i)(4 + 5i)$

Real part = $2×4 - 3×5 = 8 - 15 = -7$ Imaginary part = $2×5 + 3×4 = 10 + 12 = 22$

Answer = $-7 + 22i$ ✓

Check: FOIL gives $8 + 10i + 12i + 15i^2 = 8 + 22i - 15 = -7 + 22i$ ✓

Example 2: $(3 - 2i)(1 + 4i)$

Real part = $3×1 - (-2)×4 = 3 + 8 = 11$ Imaginary part = $3×4 + (-2)×1 = 12 - 2 = 10$

Answer = $11 + 10i$ ✓

Example 3: $(-2 + 5i)(3 - 7i)$

Real part = $(-2)×3 - 5×(-7) = -6 + 35 = 29$ Imaginary part = $(-2)×(-7) + 5×3 = 14 + 15 = 29$

Answer = $29 + 29i = 29(1+i)$ ✓

Example 4: $z \times \bar{z}$ where $z = a + ib$, $\bar{z} = a - ib$

Real part = $a×a - b×(-b) = a^2 + b^2$ Imaginary part = $a×(-b) + b×a = -ab + ab = 0$

Thus $z\bar{z} = a^2 + b^2 = |z|^2$ ✓

1.3 — Division of Complex Numbers: Paravartya Yojayet

The Paravartya Method (Transpose and Apply)

For division $\frac{a + ib}{c + id}$, the conventional method multiplies numerator and denominator by the conjugate.

Vedic Paravartya gives a direct formula:

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

Pattern Recognition

• Real part = $\frac{ac + bd}{c^2 + d^2}$
• Imaginary part = $\frac{bc - ad}{c^2 + d^2}$

Example 1: $\frac{2 + 3i}{4 + 5i}$

Denominator: $c^2 + d^2 = 4^2 + 5^2 = 16 + 25 = 41$

Real part = $\frac{2×4 + 3×5}{41} = \frac{8 + 15}{41} = \frac{23}{41}$ Imaginary part = $\frac{3×4 - 2×5}{41} = \frac{12 - 10}{41} = \frac{2}{41}$

Answer = $\frac{23}{41} + \frac{2}{41}i$ ✓

Example 2: $\frac{1 + i}{1 - i}$

Denominator: $1^2 + (-1)^2 = 1 + 1 = 2$

Real part = $\frac{1×1 + 1×(-1)}{2} = \frac{1 - 1}{2} = 0$ Imaginary part = $\frac{1×1 - 1×(-1)}{2} = \frac{1 + 1}{2} = 1$

Answer = $0 + 1i = i$ ✓

Check: $(1+i)/(1-i) = i$ (known result) ✓

Example 3: $\frac{3 - 2i}{1 + 2i}$

Denominator: $1^2 + 2^2 = 1 + 4 = 5$

Real part = $\frac{3×1 + (-2)×2}{5} = \frac{3 - 4}{5} = -\frac{1}{5}$ Imaginary part = $\frac{(-2)×1 - 3×2}{5} = \frac{-2 - 6}{5} = -\frac{8}{5}$

Answer = $-\frac{1}{5} - \frac{8}{5}i$ ✓

Paravartya Shortcut for Division

For division, the transpose rule: To divide by $c + id$, multiply numerator and denominator by $c - id$ (the transpose with sign changed).

1.4 — Modulus and Argument: Vedic Shortcuts

Definition

For $z = a + ib$:

• Modulus: $|z| = \sqrt{a^2 + b^2}$
• Argument: $\theta = \tan^{-1}\left(\frac{b}{a}\right)$ (considering quadrant)

Vedic Pattern: Modulus from Conjugate

Example: Find $|3 + 4i|$

$|3 + 4i| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$ ✓

Argument Quadrant Rules (ASTC)

Example: Find argument of $z = -1 + i$

a=-1 (negative), b=1 (positive) → Quadrant II

$\theta = \pi - \tan^{-1}(1/1) = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$ ✓

1.5 — Polar Form: Yavadunam Connection

Polar Representation

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

where $r = |z|$, $\theta = \arg(z)$

Yavadunam Insight

The word Yavadunam (whatever the extent of its deficiency) applies here as: whatever the angle θ is, it determines the position on the unit circle.

Multiplication in Polar Form (Yavadunam Pattern)

If $z_1 = r_1 e^{i\theta_1}$ and $z_2 = r_2 e^{i\theta_2}$, then:

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

Rule: Multiply moduli, add arguments.

Example: $z_1 = 2e^{i\pi/3}$, $z_2 = 3e^{i\pi/6}$

$z_1 z_2 = (2×3) e^{i(\pi/3 + \pi/6)} = 6e^{i\pi/2} = 6(\cos 90° + i\sin 90°) = 6(0 + i) = 6i$ ✓

1.6 — De Moivre's Theorem: Pattern Recognition

The Theorem

For any integer $n$:

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

Or: $(re^{i\theta})^n = r^n e^{in\theta}$

Vedic Pattern: Power = Multiply modulus, Multiply argument

Example 1: $(1 + i)^6$

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

$(1 + i)^6 = (\sqrt{2})^6 e^{i(6×\pi/4)} = 8 e^{i(3\pi/2)} = 8(\cos 270° + i\sin 270°) = 8(0 - i) = -8i$ ✓

Example 2: $(1 - i\sqrt{3})^4$

$1 - i\sqrt{3}$: $r = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2$ $\theta = \tan^{-1}(-\sqrt{3}/1) = -60° = -\pi/3$ (Quadrant IV)

$(2e^{-i\pi/3})^4 = 16 e^{-i4\pi/3} = 16(\cos 240° + i\sin 240°) = 16(-1/2 - i\sqrt{3}/2) = -8 - 8i\sqrt{3}$ ✓

1.7 — nth Roots of Unity: Cyclic Patterns

Definition

The nth roots of unity are solutions to $z^n = 1$:

$$z_k = e^{i2\pi k/n} = \cos\frac{2\pi k}{n} + i\sin\frac{2\pi k}{n}, \quad k = 0, 1, 2, ..., n-1$$

Key Properties (Vedic Pattern Recognition)

Cube Roots of Unity ($\omega$)

For $n=3$: $\omega = e^{i2\pi/3} = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$, $\omega^2 = e^{i4\pi/3} = -\frac{1}{2} - i\frac{\sqrt{3}}{2}$

Properties:

• $1 + \omega + \omega^2 = 0$
• $\omega^3 = 1$
• $\omega^2 = \bar{\omega} = \omega^{-1}$

Example: Simplify $(1 + \omega - \omega^2)^3$

Using $1 + \omega + \omega^2 = 0$ → $1 + \omega = -\omega^2$

So $1 + \omega - \omega^2 = -\omega^2 - \omega^2 = -2\omega^2$

Then $(-2\omega^2)^3 = -8\omega^6 = -8(\omega^3)^2 = -8(1)^2 = -8$ ✓

1.8 — Complex Number Equations: Shunyam Samya

Sutra 5: Shunyam Saamyasamuccaye

"If the samuccaya (whole) is the same, it is zero."

In complex equations, this helps solve equations where both sides have the same modulus or where arguments are equal.

Example 1: $|z - 1| = |z + 1|$

This means distance from (1,0) equals distance from (-1,0). The locus is the perpendicular bisector: the imaginary axis.

Solution: $\text{Re}(z) = 0$ (pure imaginary) ✓

Example 2: $|z - 2i| = |z + 2i|$

Distance from (0,2) equals distance from (0,-2). The locus is the perpendicular bisector: the real axis.

Solution: $\text{Im}(z) = 0$ (real numbers) ✓

Example 3: $\arg(z-1) = \arg(z+1)$

The arguments are equal means the vectors from (1,0) and (-1,0) to z are collinear with the origin.

This implies z lies on the real axis (Im(z)=0) but not between -1 and 1? Actually, arg equality means z is on the real axis.

Example 4: Solve $z^2 = \bar{z}$

Let $z = x + iy$. Then $x^2 - y^2 + 2ixy = x - iy$

Equating real and imaginary parts: (1) $x^2 - y^2 = x$ (2) $2xy = -y$ → $y(2x+1) = 0$

Case 1: $y=0$ → $x^2 = x$ → $x=0$ or $x=1$ → $z=0,1$ Case 2: $x = -1/2$ → $(-1/2)^2 - y^2 = -1/2$ → $1/4 - y^2 = -1/2$ → $y^2 = 3/4$ → $y = \pm\sqrt{3}/2$

Thus $z = -\frac{1}{2} \pm i\frac{\sqrt{3}}{2} = \omega, \omega^2$

Solutions: $z = 0, 1, \omega, \omega^2$ ✓

1.9 — Argand Plane: Vedic Geometric Interpretation

The Geometric View

Each complex number $z = x + iy$ corresponds to a point $(x, y)$ in the Argand plane.

Key Geometric Interpretations

Example: $|z - 3 + 4i| = 5$

Center at $(3, -4)$, radius $5$. This is the circle passing through origin? Check: $|0 - 3 + 4i| = |-3 + 4i| = 5$ ✓ So origin lies on the circle.

1.10 — JEE-Level Applications

Example 1: If $|z| = 1$, find maximum of $|z^2 + z + 1|$

Let $z = e^{i\theta}$.

$|z^2 + z + 1| = |e^{i2\theta} + e^{i\theta} + 1| = |e^{i\theta}(e^{i\theta} + 1 + e^{-i\theta})| = |e^{i\theta}| \cdot |2\cos\theta + 1|$

Since $|e^{i\theta}| = 1$, we have $|z^2 + z + 1| = |2\cos\theta + 1|$

Maximum occurs when $\cos\theta = 1$ → $2×1 + 1 = 3$

Maximum value = $3$ ✓

Example 2: Find locus of $z$ such that $\frac{z-1}{z+1}$ is purely imaginary.

Let $w = \frac{z-1}{z+1}$ be purely imaginary → $w + \bar{w} = 0$

$\frac{z-1}{z+1} + \frac{\bar{z} - 1}{\bar{z} + 1} = 0$

Cross-multiply: $(z-1)(\bar{z}+1) + (\bar{z}-1)(z+1) = 0$

Expand: $z\bar{z} + z - \bar{z} - 1 + z\bar{z} + \bar{z} - z - 1 = 0$

$2z\bar{z} - 2 = 0$ → $z\bar{z} = 1$ → $|z| = 1$

So locus is unit circle (excluding $z = -1$) ✓

PART 2: WORKED EXAMPLES

Section A: Multiplication (Urdhva)

Example 1

Question: Multiply $(3 + 2i)(1 + 4i)$.

Answer:

Real part = $3×1 - 2×4 = 3 - 8 = -5$ Imaginary part = $3×4 + 2×1 = 12 + 2 = 14$

Answer = $-5 + 14i$ ✓

Example 2

Question: Multiply $(4 - i)(2 + 3i)$.

Answer:

Real part = $4×2 - (-1)×3 = 8 + 3 = 11$ Imaginary part = $4×3 + (-1)×2 = 12 - 2 = 10$

Answer = $11 + 10i$ ✓

Section B: Division (Paravartya)

Example 3

Question: Divide $\frac{4 + 5i}{2 + 3i}$.

Answer:

Denominator: $2^2 + 3^2 = 4 + 9 = 13$

Real part = $\frac{4×2 + 5×3}{13} = \frac{8 + 15}{13} = \frac{23}{13}$ Imaginary part = $\frac{5×2 - 4×3}{13} = \frac{10 - 12}{13} = -\frac{2}{13}$

Answer = $\frac{23}{13} - \frac{2}{13}i$ ✓

Example 4

Question: Divide $\frac{1 - i}{1 + i}$.

Answer:

Denominator: $1^2 + 1^2 = 2$

Real part = $\frac{1×1 + (-1)×1}{2} = \frac{1 - 1}{2} = 0$ Imaginary part = $\frac{(-1)×1 - 1×1}{2} = \frac{-1 - 1}{2} = -1$

Answer = $-i$ ✓

Section C: Modulus and Argument

Example 5

Question: Find modulus and argument of $z = -3 + 4i$.

Answer:

$|z| = \sqrt{(-3)^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

Quadrant II (a<0, b>0) $\theta = \pi - \tan^{-1}(4/3)$

Since $\tan^{-1}(4/3) \approx 53.13°$, $\theta \approx 126.87° = 2.214 \text{ rad}$

Answer: $|z| = 5$, $\arg(z) = \pi - \tan^{-1}(4/3)$ ✓

Example 6

Question: Find modulus and argument of $z = -4 - 4i$.

Answer:

$|z| = \sqrt{(-4)^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$

Quadrant III (a<0, b<0) $\theta = -\pi + \tan^{-1}(|-4|/|-4|) = -\pi + \tan^{-1}(1) = -\pi + \pi/4 = -3\pi/4$

Alternatively: $\theta = \pi + \tan^{-1}(-4/-4) = \pi + \pi/4 = 5\pi/4$ (both valid, usually principal between -π and π is $-3\pi/4$)

Answer: $4\sqrt{2}$, $\arg = -3\pi/4$ ✓

Section D: De Moivre's Theorem

Example 7

Question: Find $(1 + i\sqrt{3})^5$.

Answer:

First convert to polar: $1 + i\sqrt{3}$ $r = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2$ $\theta = \tan^{-1}(\sqrt{3}/1) = \pi/3$ (Quadrant I)

$(2e^{i\pi/3})^5 = 32 e^{i5\pi/3} = 32(\cos 300° + i\sin 300°) = 32(1/2 - i\sqrt{3}/2) = 16 - 16i\sqrt{3}$ ✓

Example 8

Question: Find $(\cos 15° + i\sin 15°)^6$.

Answer:

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

Here $\theta = 15°$, $n = 6$

$= \cos 90° + i\sin 90° = 0 + i(1) = i$ ✓

Section E: Cube Roots of Unity

Example 9

Question: Simplify $(1 + 2\omega + \omega^2)^3$ where $\omega$ is cube root of unity.

Answer:

Using $1 + \omega + \omega^2 = 0$ → $1 + \omega^2 = -\omega$

$1 + 2\omega + \omega^2 = (1 + \omega + \omega^2) + \omega = 0 + \omega = \omega$

Then $(\omega)^3 = \omega^3 = 1$ ✓

Example 10

Question: Find the sum of all 6th roots of unity.

Answer:

Sum of all nth roots of unity = 0 for n > 1

Thus sum of 6th roots of unity = 0 ✓

Section F: Complex Equations (Shunyam Samya)

Example 11

Question: Solve $|z - 2i| = |z + 2i|$.

Answer:

Distance from (0,2) equals distance from (0,-2) → perpendicular bisector is the real axis.

Solution: $z$ is purely real ($\text{Im}(z) = 0$) ✓

Example 12

Question: Solve $z^3 = \bar{z}$.

Answer:

Let $z = re^{i\theta}$. Then $z^3 = r^3 e^{i3\theta}$, $\bar{z} = re^{-i\theta}$

Equating: $r^3 e^{i3\theta} = re^{-i\theta}$

So $r^3 = r$ → $r(r^2 - 1) = 0$ → $r = 0$ or $r = 1$

And $e^{i3\theta} = e^{-i\theta}$ → $e^{i4\theta} = 1$ → $4\theta = 2\pi k$ → $\theta = \frac{\pi k}{2}$

For $r=0$: $z=0$ For $r=1$: $\theta = 0, \pi/2, \pi, 3\pi/2$ → $z = 1, i, -1, -i$

Solutions: $z = 0, 1, -1, i, -i$ ✓

PART 3: PRACTICE EXERCISES

Exercise Set A: Multiplication & Division (20 Questions)

Perform the operation using Vedic methods.

A1. $(2 + 3i)(4 + 5i)$ A2. $(1 + i)(2 - i)$ A3. $(3 - 2i)(1 + 4i)$ A4. $(-2 + 4i)(3 - i)$ A5. $(5 + 2i)(5 - 2i)$ A6. $(2 + i)^2$ A7. $(3 - i)^2$ A8. $(1 + i)^3$ A9. $(2 + 3i)(2 - 3i)$ A10. $(4 + i)(4 - i)$ A11. $\frac{1 + 2i}{3 + 4i}$ A12. $\frac{2 + 3i}{1 - i}$ A13. $\frac{3 - i}{2 + i}$ A14. $\frac{4 + i}{2 - 3i}$ A15. $\frac{1}{i}$ A16. $\frac{1 + i}{1 - i}$ A17. $\frac{1}{1 + i}$ A18. $\frac{2 - i}{3 + 2i}$ A19. $\frac{i}{1 - i}$ A20. $\frac{1 + 2i}{1 - 2i}$

Exercise Set B: Modulus & Argument (15 Questions)

Find modulus and principal argument.

B1. $3 + 4i$ B2. $-3 + 4i$ B3. $-3 - 4i$ B4. $3 - 4i$ B5. $1 + i$ B6. $-1 + i$ B7. $-1 - i$ B8. $1 - i$ B9. $2 + 2\sqrt{3}i$ B10. $-2 + 2i$ B11. $-2 - 2\sqrt{3}i$ B12. $5i$ B13. $-5$ B14. $\sqrt{3} + i$ B15. $-\sqrt{3} - i$

Exercise Set C: De Moivre's Theorem & Powers (15 Questions)

Evaluate using De Moivre's theorem.

C1. $(1 + i)^4$ C2. $(1 - i)^6$ C3. $(\sqrt{3} + i)^4$ C4. $(-1 + i\sqrt{3})^3$ C5. $(1 + i\sqrt{3})^6$ C6. $(\cos 20° + i\sin 20°)^9$ C7. $(\cos 10° - i\sin 10°)^5$ C8. $(\frac{1}{2} + i\frac{\sqrt{3}}{2})^{12}$ C9. $(1 + i)^{20}$ C10. $(\sqrt{3} - i)^{10}$ C11. Find the cube roots of 1 C12. Find the cube roots of -1 C13. Find the fourth roots of 1 C14. Find the fourth roots of -1 C15. Find the sixth roots of 1

Exercise Set D: Cube Roots of Unity (10 Questions)

Use $\omega$ as a complex cube root of unity ($\omega \neq 1$).

D1. Simplify $1 + \omega + \omega^2$ D2. Simplify $(1 + \omega - \omega^2)^3$ D3. Simplify $(1 + 2\omega + \omega^2)^3$ D4. Find the value of $(1 - \omega + \omega^2)(1 + \omega - \omega^2)$ D5. Find the value of $(1 + \omega)^3$ D6. Find the value of $(1 - \omega)(1 - \omega^2)$ D7. Find the value of $(2 + 5\omega + 2\omega^2)^2$ D8. Show that $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a + b\omega + c\omega^2)(a + b\omega^2 + c\omega)$ D9. Find the sum of all cube roots of 1 D10. Find the product of all cube roots of 1

Exercise Set E: Complex Equations (10 Questions)

Solve for z.

E1. $|z - 1| = |z + 1|$ E2. $|z - 2| = 3$ E3. $|z - i| = |z + i|$ E4. $\arg(z) = \pi/3$ E5. $|z - 2| = |z - 4i|$ E6. $z^2 = \bar{z}$ E7. $z^2 = -1$ E8. $z^2 + z + 1 = 0$ E9. $z^4 = 16$ E10. $z^6 = 1$ (all roots)

Answer Key for Practice Exercises

Set A Answers:

A1. $-7 + 22i$
A2. $3 + i$
A3. $11 + 10i$
A4. $-2 + 14i$
A5. $29$
A6. $3 + 4i$
A7. $8 - 6i$
A8. $-2 + 2i$
A9. $13$
A10. $17$
A11. $\frac{11}{25} + \frac{2}{25}i$
A12. $-\frac{1}{2} + \frac{5}{2}i$
A13. $1 - i$
A14. $\frac{5}{13} + \frac{14}{13}i$
A15. $-i$
A16. $i$
A17. $\frac{1}{2} - \frac{1}{2}i$
A18. $\frac{4}{13} - \frac{7}{13}i$
A19. $-\frac{1}{2} + \frac{1}{2}i$
A20. $-\frac{3}{5} + \frac{4}{5}i$

Set B Answers:

B1. 5, $\tan^{-1}(4/3)$
B2. 5, $\pi - \tan^{-1}(4/3)$
B3. 5, $-\pi + \tan^{-1}(4/3)$
B4. 5, $-\tan^{-1}(4/3)$
B5. $\sqrt{2}$, $\pi/4$
B6. $\sqrt{2}$, $3\pi/4$
B7. $\sqrt{2}$, $-3\pi/4$
B8. $\sqrt{2}$, $-\pi/4$
B9. 4, $\pi/3$
B10. $2\sqrt{2}$, $3\pi/4$
B11. 4, $-2\pi/3$
B12. 5, $\pi/2$
B13. 5, $\pi$
B14. 2, $\pi/6$
B15. 2, $-5\pi/6$

Set C Answers:

C1. $-4$
C2. $8i$
C3. $-8 + 8i\sqrt{3}$
C4. $8$
C5. $64$
C6. $-1$
C7. $\cos50° - i\sin50°$
C8. $1$
C9. $-1024$
C10. $512 + 512i\sqrt{3}$
C11. $1, \omega, \omega^2$
C12. $-1, -\omega, -\omega^2$
C13. $1, -1, i, -i$
C14. $e^{iπ/4}, e^{i3π/4}, e^{i5π/4}, e^{i7π/4}$
C15. $e^{iπk/3}, k=0..5$

Set D Answers:

D1. 0
D2. $-8$
D3. 1
D4. 4
D5. $-1$
D6. 3
D7. 9
D8. Identity
D9. 0
D10. $(-1)^{2}=1$

Set E Answers:

E1. Im(z)=0 (real axis)
E2. Circle center (2,0) radius 3
E3. Re(z)=0 (imaginary axis)
E4. Ray from origin at 60°
E5. Perpendicular bisector of (2,0) and (0,4): from $|z-2|=|z-4i|$ → $(x-2)^2+y^2 = x^2+(y-4)^2$ → $-4x+4 = -8y+16$ → $x-2y+3=0$
E6. $z=0,1,\omega,\omega^2$
E7. $z=\pm i$
E8. $z=\omega,\omega^2$
E9. $z=\pm 2, \pm 2i$
E10. $z = e^{i\pi k/3}, k=0..5$

PART 5: TEACHER'S GUIDE

Common Mistakes & Corrections

Memory Aids

QUICK REFERENCE CARD

╔═══════════════════════════════════════════════════════════════════════╗
║                    MODULE 26 — COMPLEX NUMBERS: VEDIC APPROACH         ║
╠═══════════════════════════════════════════════════════════════════════╣
║                                                                       ║
║  MULTIPLICATION (Urdhva): (a+ib)(c+id) = (ac-bd) + i(ad+bc)           ║
║                                                                       ║
║  DIVISION (Paravartya): (a+ib)/(c+id) = (ac+bd)/(c²+d²) + i(bc-ad)/(c²+d²)║
║                                                                       ║
║  MODULUS: |a+ib| = √(a²+b²)                                          ║
║  ARGUMENT: θ = tan⁻¹(b/a) with quadrant adjustment (ASTC)            ║
║                                                                       ║
║  POLAR FORM: z = r(cosθ + i sinθ) = re^(iθ)                          ║
║                                                                       ║
║  DE MOIVRE: (cosθ + i sinθ)^n = cos nθ + i sin nθ                     ║
║             (re^(iθ))^n = r^n e^(inθ)                                ║
║                                                                       ║
║  CUBE ROOTS OF UNITY: 1, ω, ω², where ω = e^(i2π/3)                   ║
║  Properties: ω³=1, 1+ω+ω²=0, ω² = ω̅                                 ║
║                                                                       ║
║  nth ROOTS OF UNITY: z_k = e^(i2πk/n), k=0,...,n-1                   ║
║  Sum = 0 (for n>1), Product = (-1)^(n-1)                             ║
║                                                                       ║
║  SUTRAS: Urdhva-Tiryagbhyam (3), Paravartya (4),                     ║
║          Yavadunam (10), Shunyam Samya (5)                            ║
║                                                                       ║
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