Complex Plane and Argand Diagram for JEE Main

The Argand diagram represents complex numbers as points or position vectors in a plane, converting abstract algebra into visual geometry. JEE Main uses this representation extensively — locus questions, rotation of a complex number, distance between points, and the argument of a complex expression all become geometric on the Argand diagram. Building this visual intuition is the key to solving complex number problems fluently.

Representing Complex Numbers Geometrically

A complex number z = x + iy is the point (x, y) in the Argand plane. The modulus |z| = √(x² + y²) is its distance from the origin. The argument arg(z) = arctan(y/x) (adjusted for quadrant) is the angle from the positive real axis. The polar form z = r(cosθ + i sinθ) = re^(iθ) (Euler's form) is the most compact. Multiplication by e^(iα) rotates a point by angle α anticlockwise about the origin — a crucial geometric fact for rotation problems. For the algebraic foundation, see our complex numbers guide.

Locus Questions on the Argand Plane

Locus questions give a condition on z and ask for the shape it traces. Key results: |z − z₁| = r traces a circle centred at z₁ with radius r. |z − z₁| = |z − z₂| traces the perpendicular bisector of z₁z₂. |z − z₁|/|z − z₂| = k (k ≠ 1) traces an Apollonius circle. arg(z − z₁) = θ traces a ray from z₁ at angle θ. |z − z₁| + |z − z₂| = 2a (with a > ½|z₁ − z₂|) traces an ellipse with foci z₁, z₂. Recognise which condition you have before computing — the shape is often the answer.

Rotation and Transformation

If w = z × e^(iα), then |w| = |z| (distance from origin preserved) and arg(w) = arg(z) + α (rotated by α). This single fact solves every rotation problem. Example: to rotate vector from A to B by 90° anticlockwise about A, compute B' − A = (B − A) × e^(iπ/2) = (B − A) × i, then B' = A + i(B − A). JEE presents this as finding the fourth vertex of a square or rhombus given three vertices — always use rotation of complex numbers for these.

Modulus-Argument Form and De Moivre's Theorem

De Moivre's theorem: (cosθ + i sinθ)ⁿ = cos(nθ) + i sin(nθ). This is used to find nth roots of unity and nth powers of complex numbers. The nth roots of unity are e^(2πik/n) for k = 0, 1, ..., n−1 — they form the vertices of a regular n-gon inscribed in the unit circle. Their sum is always zero (for n ≥ 2). JEE uses cube roots (1, ω, ω²) and their properties 1 + ω + ω² = 0 and ω³ = 1 constantly in algebraic problems.

Key Inequalities

Triangle inequality: ||z₁| − |z₂|| ≤ |z₁ + z₂| ≤ |z₁| + |z₂|. These bounds are used to find the range of |z| under a constraint, and as inequality-based MCQs. The equality |z₁ + z₂| = |z₁| + |z₂| holds iff z₁ and z₂ have the same argument (collinear with origin, same side). After practising Argand diagram locus, rotation, and De Moivre problems, take a free mock test on complex numbers to assess readiness.