Complex Numbers — JEE Main & Advanced Notes

Treat complex numbers as algebra plus geometry: modulus, argument, loci, roots of unity and transformations.

Material signal: Mapped from complex-number revision assignments with straight objective and multiple-correct formats.

Priority: High Yield. Treat this as a moderate chapter inside the JEE Mathematics ladder.

• Algebraic and geometric duality: Every complex number z=x+iy is simultaneously an algebraic quantity and a point in the Argand plane. Modulus |z| is the distance from the origin; |z−a| is the distance from the fixed point a. This geometric language turns algebraic conditions into curve equations. Always plot the Argand diagram before writing equations — it immediately shows whether the locus is a circle, perpendicular bisector, ellipse or another conic.
• Principal argument and quadrant rule: The principal argument of z is the unique angle θ∈(−π, π] with tan θ = y/x and the correct quadrant sign. Q1 (x>0,y>0): θ∈(0,π/2). Q2 (x<0,y>0): θ = π−arctan|y/x|. Q3 (x<0,y<0): θ = −π+arctan|y/x|. Q4 (x>0,y<0): θ = −arctan|y/x|. arctan(y/x) alone gives values only in (−π/2,π/2) and is wrong in Q2 and Q3.
• Polar form and De Moivre theorem: Writing z = r·e^(iθ) makes multiplication trivially easy — multiply moduli and add arguments. De Moivre (cosθ+i sinθ)^n = cos(nθ)+i sin(nθ) is just repeated multiplication in polar form. Applications: (a) expand cos(nθ) or sin(nθ) as powers of cosθ and sinθ; (b) find all nth roots of any complex number.
• Finding nth roots of a complex number: To find all n roots of z = r·e^(iθ), the modulus of each root is r^(1/n) and the arguments are θ/n + 2kπ/n for k = 0,1,…,n−1. These n roots are equally spaced by 2π/n on a circle of radius r^(1/n). Cube roots are separated by 120°; square roots by 180°.
• Cube roots of unity — the fast toolkit: Non-real cube roots of 1 are ω = −1/2 + i(√3/2) and ω² = −1/2 − i(√3/2). Identities: 1+ω+ω²=0 and ω³=1. Reduce any power: ω^n = ω^(n mod 3). Expressions like a+bω+cω² are simplified using 1 = −ω−ω². Note ω and ω² are complex conjugates of each other.
• Loci in the Argand plane — systematic identification: (1) |z−a|=r: circle, centre a, radius r. (2) |z−a|=|z−b|: perpendicular bisector of segment ab. (3) |z−a|+|z−b|=2c with c>|a−b|/2: ellipse with foci a and b. (4) arg(z−a)=θ: ray from a in direction θ. Substitute z=x+iy, expand the condition, and identify the resulting conic.
• Conjugate as an algebraic tool: z·z̄ = |z|² removes i from denominators. For 1/(a+bi), multiply by (a−bi)/(a−bi) to get (a−bi)/(a²+b²). Re(z) = (z+z̄)/2 and Im(z) = (z−z̄)/(2i). These identities work purely algebraically without expanding z=x+iy.
• Rotation transformation: Multiplying z by e^(iα) rotates it anticlockwise by α about the origin. To rotate point z_P about centre z_C by angle α: z_new = z_C + (z_P−z_C)·e^(iα). For 90° anticlockwise, multiply by i. Used to prove equilateral triangles (60° rotation) and find perpendiculars (90° rotation) in JEE.

• z = x+iy (Cartesian); |z| = sqrt(x^2+y^2); arg(z): use quadrant of (x,y) — never apply arctan blindly
  Hint: do not memorise this in isolation; connect it to the definition or diagram that produces it.
• Polar/Euler form: z = r·e^(iθ) = r(cosθ + i sinθ), r = |z|, θ = arg(z)
  Hint: do not memorise this in isolation; connect it to the definition or diagram that produces it.
• Conjugate: z̄ = x−iy; z·z̄ = |z|²; z+z̄ = 2Re(z); z−z̄ = 2i·Im(z)
  Hint: do not memorise this in isolation; connect it to the definition or diagram that produces it.
• Product: |z₁z₂| = |z₁||z₂|; arg(z₁z₂) = arg(z₁)+arg(z₂) mod 2π
  Hint: do not memorise this in isolation; connect it to the definition or diagram that produces it.
• Quotient: |z₁/z₂| = |z₁|/|z₂|; arg(z₁/z₂) = arg(z₁)−arg(z₂)
  Hint: do not memorise this in isolation; connect it to the definition or diagram that produces it.
• De Moivre: (cosθ+i sinθ)^n = cos(nθ)+i sin(nθ) for all n in Z
  Hint: do not memorise this in isolation; connect it to the definition or diagram that produces it.
• nth roots of unity: w_k = e^(2πik/n), k=0…n−1; sum of all nth roots = 0; equally spaced on unit circle
  Hint: do not memorise this in isolation; connect it to the definition or diagram that produces it.
• Cube roots of unity: 1, ω=e^(2πi/3), ω²=e^(4πi/3); 1+ω+ω²=0; ω³=1; ω̄=ω²
  Hint: do not memorise this in isolation; connect it to the definition or diagram that produces it.
• Triangle inequality: |z₁+z₂| ≤ |z₁|+|z₂|; |z₁−z₂| ≥ ||z₁|−|z₂||
  Hint: do not memorise this in isolation; connect it to the definition or diagram that produces it.
• Loci: |z−a|=r → circle centre a radius r; |z−a|=|z−b| → perp bisector of segment ab
  Hint: do not memorise this in isolation; connect it to the definition or diagram that produces it.
• arg((z−z₁)/(z−z₂)) = α → arc of circle through z₁,z₂ subtending angle α
  Hint: do not memorise this in isolation; connect it to the definition or diagram that produces it.
• Rotation about origin by α: z_new = z·e^(iα); about centre c: z_new = c + (z−c)·e^(iα)
  Hint: do not memorise this in isolation; connect it to the definition or diagram that produces it.

• Using arctan(y/x) for the argument without a quadrant check: arctan gives values only in (−π/2, π/2). For z=−1+i (Q2), the correct argument is 3π/4, not −π/4. Always identify the quadrant from the signs of Re(z) and Im(z) first.
• Confusing |z₁+z₂| with |z₁|+|z₂|: the triangle inequality says |z₁+z₂| ≤ |z₁|+|z₂|. Equality holds only when z₁ and z₂ have the same argument. Moduli do not simply add for arbitrary complex numbers.
• Reducing ω^n without computing n mod 3: ω^100 — 100=33×3+1 so ω^100=ω. Skipping this step leads to wrong expressions in evaluation problems.
• Squaring both sides of a modulus equation as if z is real: |z−a|² = (z−a)(z̄−ā), not (z−a)². Always expand using z·z̄ = |z|².
• Writing arg(z₁/z₂) = arg(z₁)/arg(z₂): arguments subtract under division. The correct identity is arg(z₁/z₂) = arg(z₁) − arg(z₂). Division of arguments is meaningless.

For JEE Main, aim for fast recognition and clean substitution. Finish the first pass of Complex Numbers questions in 60–90 seconds each. Prioritise standard formulas, short sign/domain checks and option elimination only after the setup is correct.

For JEE Advanced, expect combined conditions, hidden domains, multi-correct traps and integer-style answers. Build the solution from definitions, not memorised tricks. When a parameter appears, solve the general case and then filter using restrictions.

• Draw the Argand point first. Translate |z−a|=r immediately as circle, centre a, radius r.
• Argument: find quadrant from signs of Re and Im; use arctan|Im/Re|; apply sign correction for that quadrant.
• nth roots: modulus = r^(1/n); arguments = θ/n + 2kπ/n for k=0,1,…,n−1.
• Cube root powers: reduce exponent mod 3. Remainder 0→1, 1→ω, 2→ω². Use 1+ω+ω²=0 to collapse sums.
• Rotation: multiply (z − centre) by e^(iα), then add centre back. For 90° anticlockwise, multiply by i.