Complex Numbers

Complex numbers unify algebra, trigonometry and geometry. Advanced exploits the geometric meaning of |z| and arg(z), rotation via multiplication by e^(iθ), roots of unity, and loci in the Argand plane. Fluency in the polar/Euler form turns hard geometry into one-line algebra.

Key results

• Forms: z = x + iy = r(cosθ + i sinθ) = re^(iθ), r = |z| = √(x²+y²), θ = arg z.
• De Moivre: (cosθ + i sinθ)ⁿ = cos nθ + i sin nθ; nth roots of unity are e^(2πik/n), summing to 0 and forming a regular n-gon.
• Rotation: multiplying by e^(iα) rotates a vector by α; (z2 − z1)/(z3 − z1) encodes angle and ratio at z1.
• Triangle inequality: ||z1| − |z2|| ≤ |z1 ± z2| ≤ |z1| + |z2|.
• Loci: |z − a| = r (circle), |z − a| = |z − b| (perpendicular bisector), arg((z−a)/(z−b)) = const (arc).

Object	Complex condition
Circle (centre a)
Perp. bisector
Collinearity	(z2−z1)/(z3−z1) real
Rotation by α	multiply by e^(iα)

Exam Tricks & Tips

• 🎯 Multiplication by e^(iθ) is a rotation — use it to rotate points/vectors and to prove geometry results without coordinates.
• 🎯 The n nth-roots of unity sum to zero and any power sum 1+ω^k+ω^(2k)+... = 0 unless k is a multiple of n — a staple for cube-root-of-unity (ω) identities (1+ω+ω²=0).
• 🎯 |z1 + z2| = |z1| + |z2| iff z1 and z2 have the same argument (equality in the triangle inequality) — a common discriminator.
• 🎯 arg((z − z1)/(z − z2)) = constant is an arc of a circle through z1 and z2 (90° gives the circle on z1z2 as diameter).
• 🎯 For rotation of z2 about z1 by angle θ: z' = z1 + (z2 − z1)e^(iθ) — memorise for square/equilateral-triangle constructions.
• ❌ Common mistake: taking arg of a product as the plain sum without adjusting by ±2π to keep it in the principal range (−π, π].

Expected exam pattern

2 questions likely: a roots-of-unity/De Moivre computation and a locus/geometry (rotation, arg condition) problem, often multiple-correct. High-value and recurring.

Quick recap

Use z = re^(iθ); multiply by e^(iα) to rotate. De Moivre and roots of unity (sum 0, 1+ω+ω²=0). Loci: |z−a|=r circle, |z−a|=|z−b| bisector, arg ratio = arc. Triangle inequality equality needs equal arguments.

Cover the answer, recall, then check. 12 cards on complex numbers for JEE Advanced.

Q1. Polar/Euler form of a complex number.
A1. z = r(cosθ + i sinθ) = re^(iθ), r = |z|, θ = arg z.

Q2. De Moivre's theorem.
A2. (cosθ + i sinθ)ⁿ = cos nθ + i sin nθ.

Q3. Sum of all n nth-roots of unity.
A3. Zero (they form a regular n-gon centred at the origin).

Q4. Value of 1 + ω + ω² for the cube roots of unity.
A4. 0 (and ω³ = 1).

Q5. Geometric effect of multiplying z by e^(iα).
A5. Rotation of the vector z by angle α about the origin.

Q6. Locus of |z − a| = r.
A6. A circle of radius r centred at a.

Q7. Locus of |z − a| = |z − b|.
A7. The perpendicular bisector of the segment joining a and b.

Q8. Rotation formula for z2 about z1 by angle θ.
A8. z' = z1 + (z2 − z1)e^(iθ).

Q9. Condition for equality |z1 + z2| = |z1| + |z2|.
A9. z1 and z2 have the same argument (are collinear with origin, same direction).

Q10. Condition for three points z1, z2, z3 to be collinear.
A10. (z2 − z1)/(z3 − z1) is real.

Q11. What does arg((z − z1)/(z − z2)) = π/2 represent?
A11. The circle with z1z2 as diameter (the arc where the segment subtends 90°).

Q12. Modulus and argument of a product z1z2.
A12. |z1z2| = |z1||z2|; arg = arg z1 + arg z2 (adjusted to the principal range).

Complex numbers turn hard trigonometry and geometry into clean algebra. JEE Advanced treats them as a geometric tool — rotations, loci, regular polygons — far beyond the "i² = −1" basics, rewarding students who fluently switch between Cartesian, polar and exponential forms.

Core concept: a complex number z = x + iy = r(cosθ + i sinθ) = re^(iθ) is a point (or vector) in the plane; multiplication rotates and scales, which is the source of its geometric power.

Core ideas

Beginner — forms and modulus-argument

z = x + iy has modulus |z| = √(x²+y²) and argument θ = arg(z). Polar form z = r(cosθ + isinθ); Euler e^(iθ) = cosθ + isinθ. Conjugate z̄ = x − iy, with z·z̄ = |z|². Key properties: |z₁z₂| = |z₁||z₂|, arg(z₁z₂) = argz₁ + argz₂ — multiplication adds arguments (rotates).

Intermediate — De Moivre and roots of unity

De Moivre: (cosθ + isinθ)ⁿ = cos nθ + isin nθ, the engine for powers and roots. The nth roots of unity are e^(2πik/n), k = 0…n−1 — n equally spaced points on the unit circle. They sum to zero (1 + ω + ω² + … = 0) and their product is ±1. The cube roots 1, ω, ω² (with ω = e^(2πi/3)) satisfy 1 + ω + ω² = 0 and ω³ = 1 — an endlessly useful pair of identities for simplifying expressions.

Advanced — geometry and loci

Complex numbers encode plane geometry:

• Rotation: to rotate z about z₀ by angle α, use (z − z₀) → (z − z₀)e^(iα).
• Section/collinearity/rotation conditions come from arguments of ratios: three points are collinear iff (z₃−z₁)/(z₂−z₁) is real; perpendicular iff it is purely imaginary.
• Loci: |z − a| = |z − b| is the perpendicular bisector; |z − a| = k a circle; arg((z−a)/(z−b)) = constant an arc of a circle. |z − z₁| + |z − z₂| = constant is an ellipse.
  Recognising the locus from an equation, and using rotation to solve polygon/triangle problems, is the JEE-Advanced heart of the topic.

Worked example

If z₁, z₂, z₃ are the vertices of an equilateral triangle, prove z₁² + z₂² + z₃² = z₁z₂ + z₂z₃ + z₃z₁. Rotating one side onto the next by 60° gives (z₂ − z₁) = (z₃ − z₁)e^(±iπ/3), and manipulating the rotation relations for all vertices leads to the symmetric condition z₁² + z₂² + z₃² = z₁z₂ + z₂z₃ + z₃z₁. The insight — an equilateral triangle is a 60° rotation, expressible in one complex equation — is exactly the geometry-as-algebra move JEE rewards.

How JEE Advanced tests this

Loci identification (|z−a|/|z−b| ratios, argument conditions → circles, lines, ellipses); rotation problems (rotate a vertex, find a square's/triangle's fourth point); roots-of-unity summation identities and polygon geometry; De Moivre powers and nth roots; and inequalities like |z₁ + z₂| ≤ |z₁| + |z₂| (triangle inequality) with equality conditions.

Exam tricks & shortcuts

• Multiplication rotates: multiply by e^(iα) to rotate by α.
• 1 + ω + ω² = 0, ω³ = 1 — memorise for cube-root simplifications.
• nth roots of unity sum to 0 and lie equally spaced on the unit circle.
• Collinear iff ratio real; perpendicular iff ratio imaginary.

Key formulas

• z = a + ib; |z| = √(a²+b²); z̄ = a − ib; z·z̄ = |z|².
• Polar/Euler: z = r(cosθ + i sinθ) = re^(iθ), θ = arg z.
• De Moivre: (cosθ + i sinθ)ⁿ = cos nθ + i sin nθ.
• nth roots of unity: e^(2πik/n), k = 0…n−1; sum = 0.
• Properties: |z₁z₂| = |z₁||z₂|; arg(z₁z₂) = argz₁ + argz₂; |z₁+z₂| ≤ |z₁|+|z₂|.

Polar form makes multiplication, powers and roots easy; the triangle inequality bounds sums of moduli.

Worked Example

Problem: Let ω be a complex (non-real) cube root of unity, so 1 + ω + ω² = 0 and ω³ = 1. Evaluate (a) (1 + ω)⁷ and (b) (1 − ω + ω²)(1 + ω − ω²).

Solution:
Use the key relation 1 + ω + ω² = 0, which gives 1 + ω = −ω² and 1 + ω² = −ω.

(a) (1 + ω)⁷ = (−ω²)⁷ = −ω¹⁴.
Reduce the exponent modulo 3 (since ω³ = 1): 14 = 3(4) + 2, so ω¹⁴ = ω².
Therefore (1 + ω)⁷ = −ω².

(b) Simplify each factor:
1 − ω + ω² = (1 + ω²) − ω = −ω − ω = −2ω.
1 + ω − ω² = (1 + ω) − ω² = −ω² − ω² = −2ω².
Product = (−2ω)(−2ω²) = 4ω³ = 4(1) = 4.

Answer: (a) (1 + ω)⁷ = −ω²; (b) (1 − ω + ω²)(1 + ω − ω²) = 4.