# Triangle Inequality in Complex Numbers

> Triangle inequality in complex numbers for JEE Mathematics: equality conditions, reverse inequality, min-max of |z − a| ± |z − b| and the rhombus proof.

- Canonical URL: https://prepelephant.com/topics/jee/mathematics/triangle-inequality-complex
- 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: "Triangle Inequality in Complex Numbers", PrepElephant, https://prepelephant.com/topics/jee/mathematics/triangle-inequality-complex

## Direct answer

Equality holds in |z1 + z2| ≤ |z1| + |z2| exactly when z1 and z2 point the same way — same argument, one possibly zero; the reverse inequality |z1 + z2| ≥ ||z1| − |z2||| is tight when they point oppositely. The inequality earns its keep through optimisation: |z − a| + |z − b| attains its minimum |a − b| on the segment joining a and b, any larger constant tracing an ellipse; |z − a| − |z − b| always lies between ±|a − b|, attaining the endpoints on rays beyond the foci. And the structural gem: |z1| = |z2| forces z1 + z2 and z1 − z2 perpendicular — the rhombus diagonals theorem wearing complex clothing.

## What you must remember

- **Forward inequality:** |z1 + z2| ≤ |z1| + |z2|, equality iff z1 and z2 have the same argument (or one is zero).
- **Reverse inequality:** |z1 + z2| ≥ ||z1| − |z2||, equality iff opposite arguments; both bounds are sharp, not decorative.
- **Distance reading:** |z1 − z2| is the distance between the points — the inequality is the geometric fact that one side of a triangle never exceeds the sum, never falls below the difference, of the other two.
- **Minimum of a sum:** min over z of |z − a| + |z − b| equals |a − b|, achieved on the segment; any larger constant value traces an ellipse with foci a and b.
- **Bounded difference:** |z − a| − |z − b| lies in [−|a − b|, |a − b|], with equality on rays beyond the foci — hyperbola connections.
- **Rhombus theorem:** |z1| = |z2| ≠ 0 forces z1 + z2 ⊥ z1 − z2, because (z1 + z2)·(z1 − z2) = |z1|² − |z2|² = 0 in vector form.
- **Expansion identity:** |z1 + z2|² = |z1|² + |z2|² + 2Re(z1z̄2) — the dot-product machinery behind every proof here.

## A rhombus proof by inequality

Suppose |z1| = |z2| = r; prove z1 + z2 and z1 − z2 are perpendicular. Treat the complex numbers as vectors: the scalar product (z1 + z2)·(z1 − z2) expands to |z1|² + z2·z1 − z1·z2 − |z2|², where the mixed terms cancel because a·b = b·a, leaving |z1|² − |z2|² = r² − r² = 0. Perpendicular, in one line — and the geometry explains why: z1 and z2 as adjacent sides build a rhombus (all sides r), z1 + z2 and z1 − z2 are its diagonals, and a rhombus's diagonals always cross at right angles. The complex statement is that (z1 + z2)/(z1 − z2) is purely imaginary whenever |z1| = |z2| — the recognition behind countless "show the quotient is imaginary" items. Now the optimisation face: minimise |z − 1| + |z + 1|. The two foci are 1 and −1, distance 2 apart, so the triangle inequality gives |z − 1| + |z + 1| ≥ |(z − 1) − (z + 1)| = 2, with equality when z lies on the segment — the minimum is 2, attained all along the real interval [−1, 1].

## Optimisation with the inequality

JEE Main asks the min/max family directly: minimum of |z − 2| + |z + 2| is 4; maximum of |z − 3| − |z + 3| is 6 (on the ray beyond 3); minimum of |z − 1| − |z + 1| is 0, attained on the perpendicular bisector. The distractors swap sum for difference, and the equality cases separate candidates. JEE Advanced prefers the structural results: the rhombus perpendicularity above, its converse (if z1 + z2 ⊥ z1 − z2 then the moduli match), and locus questions — the set where |z − a| + |z − b| = constant greater than |a − b| is an ellipse, the constant-difference set a hyperbola, and the inequality is what certifies the constant's admissible range. The habit that holds marks: write both inequalities around the expression and ask which bound is achievable and where.

## Frequently asked questions

### When does equality hold in |z1 + z2| ≤ |z1| + |z2|?

When z1 and z2 share the same argument (collinear vectors pointing the same way), or when one of them is zero.

### What is the reverse triangle inequality?

|z1 + z2| ≥ ||z1| − |z2||, with equality when the two vectors point in opposite directions.

### What is the minimum of |z − a| + |z − b| over all z?

|a − b|, attained exactly on the segment joining a and b; any larger constant traces an ellipse.

### Why are z1 + z2 and z1 − z2 perpendicular when |z1| = |z2|?

They are the diagonals of the rhombus built on the equal vectors z1, z2 — diagonals of a rhombus cross at right angles, as the dot product |z1|² − |z2|² = 0 confirms.

### How do these inequalities connect to conics?

The ellipse is the constant-sum locus (sum above |a − b|), the hyperbola the constant-difference locus — the inequalities certify which constants are geometrically possible.