Complex Numbers

Complex Numbers

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

A complex number has the form z = a + ib, where a, b are real, i is the imaginary unit with i² = −1, a = Re(z), b = Im(z). The modulus is |z| = √(a² + b²) and the argument is arg(z) = tan⁻¹(b/a) with quadrant correction (range (−π, π] or [0, 2π)). The conjugate is z̄ = a − ib, and the identity z · z̄ = |z|² = a² + b² turns division into multiplication. De Moivre’s theorem states (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ, the workhorse for powering and rooting. In CUET UG, expect 1–2 short questions: convert to polar form, find square roots, or simplify a power — keep i² = −1 and quadrant rules at your fingertips.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Algebra and the Argand Plane

A complex number z = a + ib is plotted on the Argand plane with the real axis as x and the imaginary axis as y. Addition is vector addition: (a + ib) + (c + id) = (a + c) + i(b + d). Multiplication expands with i² = −1: (a + ib)(c + id) = (ac − bd) + i(ad + bc). Division uses the conjugate trick:

z₁ / z₂ = (z₁ · z̄₂) / |z₂|²

Modulus and Conjugate Properties

• |z|² = z · z̄ = a² + b² (never z²)
• |z₁ z₂| = |z₁| · |z₂| and |z₁ / z₂| = |z₁| / |z₂|
• Triangle inequality: |z₁ + z₂| ≤ |z₁| + |z₂|, with equality only when arg(z₁) = arg(z₂)
• z + z̄ = 2 Re(z) and z − z̄ = 2i Im(z)

Polar Form and De Moivre

Write z = r(cos θ + i sin θ) = r e^(iθ) with r = |z| and θ = arg(z). Then:

• Multiplication: r₁ r₂ [cos(θ₁ + θ₂) + i sin(θ₁ + θ₂)]
• Power (De Moivre): zⁿ = rⁿ (cos nθ + i sin nθ)
• n-th roots: z^(1/n) = r^(1/n) [cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n)], k = 0, 1, …, n−1

Typical CUET Question Patterns

1. Find modulus and argument of (1 + i)/(1 − i).
2. If z² = 5 + 12i, find z. (Compare real and imaginary parts after setting z = a + ib.)
3. Express (1 + i√3)⁸ in polar form and simplify using De Moivre.

Key Facts

Form	Expression	Use
Rectangular	a + ib	Algebra
Modulus-Argument	r(cos θ + i sin θ)	Geometry, powers
Euler	r e^(iθ)	Calculus, series

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Square Root of a Complex Number

To find √(a + ib), set √(a + ib) = x + iy with x, y ≥ 0. Squaring gives x² − y² = a and 2xy = b, solved via x² = (√(a² + b²) + a)/2 and y² = (√(a² + b²) − a)/2. The sign of b decides the sign of y. This is the standard CUET 2-mark question.

Quadratics with Negative Discriminant

When the discriminant D = b² − 4ac < 0, roots are complex conjugates: x = (−b ± i√|D|) / (2a). The product of roots equals c/a (real) and the sum equals −b/a (real) — a fact often tested in CUET General Test.

Common Mistakes

• Writing |z|² = z²: always use z · z̄ instead; z² = (a² − b²) + i(2ab).
• Ignoring the quadrant: arg(z) for z = −1 − i is −3π/4, not π/4.
• Forgetting the rⁿ factor in De Moivre — the modulus r must be raised too, not just the angle.
• Treating |z₁ + z₂| as equal to |z₁| + |z₂|; equality only holds for collinear vectors with the same direction.
• Dividing without conjugating: (1 + i)/(1 − i) needs the conjugate 1 + i on top and bottom before simplification.

Connection to Adjacent Topics

De Moivre links directly to trigonometry (multiple-angle formulas) and binomial theorem (via (cos θ + i sin θ)ⁿ expansion). The Euler form e^(iθ) = cos θ + i sin θ ties complex numbers to exponential series and surfaces again in CUET’s calculus-light integrals.

Worked Micro-Example

Find the modulus and argument of z = (1 + i√3) / (1 + i).

1. Multiply top and bottom by 1 − i: z = (1 + i√3)(1 − i) / ((1 + i)(1 − i)) = ((1 + √3) + i(√3 − 1)) / 2.
2. Modulus: |z| = (1/2)·√((1 + √3)² + (√3 − 1)²) = (1/2)·√(1 + 2√3 + 3 + 3 − 2√3 + 1) = (1/2)·√8 = √2.
3. Argument: tan θ = (√3 − 1)/(√3 + 1) = 2 − √3 ⇒ θ = π/12 (since tan 15° = 2 − √3).

Practice Prompts

1. If z = 2(cos 40° + i sin 40°), compute z⁵ in rectangular form and state its modulus.
2. Solve z² + (2 − 3i)z + (5 − i) = 0 for z, given that one root is −1 + 2i.

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