# Geometry of Complex Numbers in the Argand Plane

> Argand plane geometry for JEE Mathematics: equation of circle and line in z, rotation formula, arg conditions, ellipse and Apollonius loci.

- Canonical URL: https://prepelephant.com/topics/jee/mathematics/geometry-complex-argand
- Exam / course: JEE · Subject: Mathematics
- Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team
- First published: 2026-10-02
- Last updated: 2026-10-02
- How to cite: "Geometry of Complex Numbers in the Argand Plane", PrepElephant, https://prepelephant.com/topics/jee/mathematics/geometry-complex-argand

## Direct answer

Every complex number z = x + iy is a point (x, y) in the Argand plane, and the standard loci translate directly: |z − z₀| = r is a circle centred at z₀ with radius r; |z − z₁| = |z − z₂| is the perpendicular bisector of the segment joining z₁ and z₂; |z − z₁|/|z − z₂| = k is an Apollonius circle for k ≠ 1 and a bisector for k = 1. Arguments give angular conditions: arg((z − z₁)/(z₂ − z₁)) = ±π/2 says the segment from z₁ to z₂ subtends a right angle at z, so z lies on the circle with diameter z₁z₂. Rotation by angle θ about the point z₀ is multiplication: z − z₀ = e^(iθ)(z' − z₀) is the form every rotation question eventually uses.

## What you must remember

- **Circle and line:** |z − z₀| = r is a circle; the general equation |z|² = z·z̄ + z̄·z + c (real coefficients) or |z − z₁| = |z − z₂| gives lines and bisectors.
- **Apollonius:** |z − z₁| = k|z − z₂| (k > 0, k ≠ 1) is a circle; when k = 1 it degenerates to the perpendicular bisector — the degeneration is itself a question.
- **Rotation formula:** rotating z about z₀ by θ: (z − z₀) → e^(iθ)(z − z₀); multiplication by i alone is a 90° anticlockwise turn about the origin.
- **Right-angle condition:** arg((z − z₁)/(z − z₂)) = π/2 means the angle at z in triangle z₁zz₂ is right, so z lies on the circle with diameter z₁z₂ (angle in a semicircle).
- **Collinearity and perpendicularity:** z₁, z₂, z₃ are collinear when (z₃ − z₁)/(z₂ − z₁) is real; the segments are perpendicular when the same ratio is purely imaginary.
- **Ellipse and hyperbola:** |z − z₁| + |z − z₂| = 2a is an ellipse (2a > |z₁ − z₂|); the difference of distances constant gives a hyperbola — loci Advanced likes to hide inside distance language.
- **Centroid and conjugate symmetry:** the centroid of triangle with vertices z₁, z₂, z₃ is (z₁ + z₂ + z₃)/3; reflection in the x-axis is conjugation, in the y-axis is −z̄, in the origin is −z.

## Reading an Argand diagram like equations

Suppose z moves so that |z − 1| = |z + i|. Squaring both sides: (x − 1)² + y² = x² + (y + 1)², which collapses to −2x = 2y, i.e. y = −x — the perpendicular bisector of the segment from 1 to −i, exactly as the geometry promised. Now sharpen it: |z − 1| = 2|z + i| gives, after squaring, x² + y² − 2x + 1 = 4(x² + y² + 2y + 1), so 3x² + 3y² + 2x + 8y + 3 = 0, a genuine circle whose centre (−1/3, −4/3) and radius can be read by completing squares. The algebraic route (square the modulus) and the geometric diagnosis must agree; diagnosing the locus type first saves time because half the demanded information comes free.

Angular conditions run the same way: arg((z − 1)/(z + 1)) = π/4 places z on an arc of the circle centred at i with radius √2 — the argument fixes which side of the chord, distinguishing an arc from the full circle.

## Main versus Advanced on Argand

JEE Main tests identification: state the locus of |z − 2| + |z + 2| = 6 (ellipse with foci ±2, a = 3, so b² = 5) — recognition plus one computation. JEE Advanced composes: rotate a point about another point, then impose a locus condition, or combine a modulus equation with an argument equation and count solutions. The recurring trap is the sign of rotation — multiplying by e^(−iθ) when the question turns clockwise — and the second is treating |z − z₁|/|z − z₂| = k as a circle when k = 1, where it flattens into a line.

## Frequently asked questions

### What curve is |z − z₁| + |z − z₂| = 2a?

An ellipse with foci at z₁ and z₂, valid when 2a > |z₁ − z₂|; the corresponding difference-of-distances locus is a hyperbola.

### What does arg((z − z₁)/(z₂ − z₁)) = π/2 mean geometrically?

The segment from z₁ to z₂ subtends a right angle at z, placing z on the circle with diameter z₁z₂.

### How do you rotate a complex number about a point other than the origin?

Use z_new = z₀ + e^(iθ)(z − z₀): translate the centre to the origin, rotate, translate back.

### When is |z − z₁| = k|z − z₂| a straight line rather than a circle?

Only when k = 1: the Apollonius ratio degenerates to the perpendicular bisector of z₁z₂.

### How do you test collinearity of three complex points?

(z₃ − z₁)/(z₂ − z₁) must be purely real; purely imaginary instead means the segments z₁z₃ and z₁z₂ are perpendicular.