JC1 H2 Maths Bukit Timah | Argand Diagram Distance

Did you know? The expression |z₁ − z₂| can tell you exactly how far apart two points are on a diagram—even though z₁ and z₂ may contain the imaginary number i. The answer is an ordinary non-negative real distance, not an imaginary number. In H2 Mathematics, that small insight turns some intimidating complex-number questions into familiar Pythagoras.

For students looking for a JC1 H2 Mathematics Tutor Bukit Timah, Argand diagrams are a natural bridge from coordinate geometry to complex numbers. Each complex number z = x + yi represents a point with horizontal coordinate x and vertical coordinate y. The Singapore–Cambridge H2 Mathematics 9758 syllabus includes the modulus, argument and conjugate of a complex number together with Argand diagrams. A learner who already understands distances between points can use that foundation rather than memorising unrelated complex-number tricks.

This guide explains why the modulus of a difference equals a distance, how to handle real and imaginary parts carefully, and how a condition such as |z − (2 + i)| = 3 describes a geometric circle. There are worked examples, sign checks, equality tests, parent-friendly questions and an independent practice section. The focus is narrow by design: the meaning of complex modulus as geometry, not every operation on complex numbers.

The quick answer: read real and imaginary parts as coordinates

If z = x + yi, with x and y real, then its modulus is |z| = √(x² + y²). On an Argand diagram this is the distance from the origin (0, 0) to the point (x, y).

For two numbers z₁ = a + bi and z₂ = c + di, their difference is z₁ − z₂ = (a − c) + (b − d)i. Therefore

|z₁ − z₂| = √[(a − c)² + (b − d)²].

This is exactly the ordinary Cartesian distance formula between points (a, b) and (c, d).

A complex number z = x + yi is represented by the point (x, y).
The real part, x, gives its horizontal coordinate; the imaginary part, y, gives its vertical coordinate.
|z| measures distance from the origin and is never negative.
|z − w| measures distance from point w to point z.
The equation |z − w| = r describes a circle centred at w when r is positive.
The inequality |z − w| ≤ r describes the filled disk of radius r around w when r is non-negative.
The condition |z − w| = 0 gives only z = w.

The formula becomes easy to remember once you see the two coordinate differences. The squares remove direction signs because distance is a magnitude.

Why an imaginary part can contribute to a real distance

The symbol i satisfies i² = −1. That does not mean the point y on an Argand diagram is unreal or that its vertical coordinate cannot be measured. We are drawing the real number x horizontally and the real coefficient y vertically. The notation yi marks a perpendicular component in this two-dimensional representation.

For example, z = 3 + 4i is plotted at (3, 4). The line from the origin to this point is the hypotenuse of a right triangle with side lengths three and four. Its length is √(3² + 4²) = 5. Hence |3 + 4i| = 5.

The result is not 3 + 4, and it is not √(3² − 4²). It is the Pythagorean length of a two-dimensional vector. The i indicates the second coordinate, while the modulus operator returns a non-negative real number.

If a student asks “How can an imaginary value have a real length?”, separate the *number* from its *coordinate representation*. The real and imaginary parts are real coordinates; the geometric length uses the squares of those real coordinates.

Worked example 1: distance between two complex numbers

Let z₁ = 1 + 2i and z₂ = 4 − 2i. Find the distance between their points.

First subtract: z₁ − z₂ = (1 − 4) + [2 − (−2)]i = −3 + 4i.

Next calculate its modulus: |−3 + 4i| = √[(-3)² + 4²] = √(9 + 16) = 5.

Now verify with ordinary coordinates. The first point is (1, 2); the second is (4, −2). Moving from the first to the second changes x by three and y by minus four. The distance is √(3² + (−4)²) = 5.

Both routes agree. The difference z₁ − z₂ points from z₂ towards z₁, whereas z₂ − z₁ points in the opposite direction. Their moduli are equal, so distance does not depend on the direction in which we subtract.

Why subtraction order does not change distance

We have z₂ − z₁ = −(z₁ − z₂). For any complex number u, |−u| = |u|, since reversing a vector does not change its length.

Therefore |z₁ − z₂| = |z₂ − z₁|. This is a useful error check when a student worries about choosing the right order of the two named points.

But note what the equality does not say. The complex differences themselves are generally unequal. In our example, z₁ − z₂ = −3 + 4i, whereas z₂ − z₁ = 3 − 4i. Equal distance is not equal displacement.

A tutor can ask learners to point at the arrows in opposite directions and explain why the arrows have equal lengths. That develops geometric intuition before more algebra enters.

Worked example 2: a complex modulus from the origin

Take z = −5 + 12i. Its position on the Argand diagram is (−5, 12). The horizontal direction is negative, but distance from the origin is non-negative.

Calculate |z| = √[(-5)² + 12²] = √(25 + 144) = 13.

If the student writes −13 because the real part is negative, they have confused a coordinate sign with a length. The modulus cannot be negative.

We can also check z̄, the complex conjugate of z. It is −5 − 12i, which reflects the point across the real axis. Its distance from the origin remains 13, so |z̄| = |z|.

This is one reason complex conjugation is useful. It reverses the imaginary coordinate, but does not alter the modulus. In the more general identity z z̄ = |z|², the multiplication produces the non-negative real number x² + y².

Why multiplying by the conjugate reveals the modulus

Write z = x + yi and z̄ = x − yi. Multiply:

z z̄ = (x + yi)(x − yi).

The middle terms cancel. Since i² = −1, the last product is −y²i² = +y². Thus z z̄ = x² + y² = |z|².

For z = 3 + 4i, the conjugate is 3 − 4i. Their product is 9 + 16 = 25, and the positive square root gives modulus five.

This algebra confirms the geometric picture. The modulus is not a mysterious extra rule for complex numbers; it is the same Euclidean length encoded by conjugation and Pythagoras.

Do not write |z| = z z̄. The correct identity relates the *square* of the modulus to that product. For a non-zero number, forgetting the square root gives the square of the distance, not the distance itself.

Worked example 3: the circle centred at a complex number

Consider the condition |z − (2 + i)| = 3.

Let z = x + yi. Then z − (2 + i) = (x − 2) + (y − 1)i. Take the modulus:

√[(x − 2)² + (y − 1)²] = 3.

Square both sides, noting both sides are non-negative, to obtain (x − 2)² + (y − 1)² = 9.

That is a circle with centre (2, 1) and radius three on the Argand diagram. The number 2 + i specifies its centre; the real number three specifies the distance every point z must have from that centre.

We can test a point. Let z = 5 + i, corresponding to (5, 1). Its distance from 2 + i is |(5 + i) − (2 + i)| = |3| = 3, so it lies on the circle.

The point z = 2 + i is the centre, not a point on the circumference. Its distance from itself is zero, which is not three. This quick check catches a common drawing error.

Why the brackets matter

The expression |z − (2 + i)| must subtract the entire complex centre. A student who writes z − 2 + i has accidentally changed the centre to 2 − i, because the bracketed imaginary term was not subtracted.

The algebra becomes clear if the learner writes [x + yi] − [2 + i] = (x − 2) + (y − 1)i before taking the modulus.

That single bracket prevents a reflection of the intended circle across the real axis.

Worked example 4: a filled disk rather than a circumference

What if the condition is |z − (2 + i)| ≤ 3?

The same centre and radius are involved, but the inequality allows every point whose distance from the centre is at most three. In Cartesian form,

(x − 2)² + (y − 1)² ≤ 9.

The solution set is the closed disk including its circular boundary. For a strict inequality < 3, only the interior points would be included; the circumference itself would be excluded.

A picture helps here. The equality describes a one-dimensional boundary curve in the plane, whereas the inequality describes a whole region.

An exam question may ask a student to shade a locus or describe where a complex number can lie. Choosing the correct boundary inclusion is therefore as important as calculating the radius.

What if the radius is negative or zero?

The modulus is never negative. Therefore |z − w| = −2 has no solutions, because a non-negative distance cannot equal a negative number.

The condition |z − w| = 0 has exactly one solution: z = w. All other points are strictly farther than zero away from w.

For |z − w| ≤ 0, the solution is again just the point w. For |z − w| < 0, there are no points. These edge cases expose why blindly drawing a “circle of radius −2” makes no geometric sense.

Before sketching any modulus locus, check that the proposed radius is compatible with a distance.

Worked example 5: two equal distances form a straight-line locus

Consider |z − (1 + i)| = |z − (5 + i)|. The complex points 1 + i and 5 + i correspond to A = (1, 1) and B = (5, 1).

We want every point z that is equally far from A and B. In ordinary geometry, these points form the perpendicular bisector of AB.

The midpoint of AB is (3, 1). Since AB is horizontal, its perpendicular bisector is the vertical line x = 3.

Verify algebraically. Let z = x + yi and square both distances:

(x − 1)² + (y − 1)² = (x − 5)² + (y − 1)².

The matching (y − 1)² terms cancel. Expanding gives x² − 2x + 1 = x² − 10x + 25. Hence 8x = 24 and x = 3.

There is no restriction on y. Every point on that vertical line is equally far from the two complex points. The geometry predicted the answer before the algebra confirmed it.

Why equal distance does not mean the points are equal

A learner might interpret |z − a| = |z − b| as z − a = z − b and conclude a = b. That step is invalid. Equal moduli do not imply equal complex numbers. Two arrows can have different directions while having the same length.

For example, |3 + 4i| = 5 and |−3 + 4i| = 5, but the complex numbers are not equal.

The same issue occurs in real numbers: |3| = |−3|, yet 3 ≠ −3. In the complex plane, there are infinitely many different complex numbers with the same modulus. They lie on the same circle centred at the origin.

This is an important distinction between an equality of numbers and an equality of distances.

Worked example 6: distance between z and its conjugate

Let z = x + yi and z̄ = x − yi. The two points have the same real coordinate but opposite imaginary coordinates. Their displacement is

z − z̄ = 2yi.

Therefore |z − z̄| = 2|y|.

Geometrically, the points are vertically aligned, at heights y and −y. The vertical distance between them is 2|y|. If z = 3 + 4i, its conjugate is 3 − 4i and the distance between them is eight units.

This gives a direct geometric meaning to conjugation: it is a reflection across the real axis, and the distance from a point to its reflected image equals twice its absolute vertical coordinate.

Do not confuse this with the fact |z| = |z̄|. Those two numbers have equal distance from the origin, but may be separated from each other.

The triangle inequality as a distance check

For any complex numbers u and v, the triangle inequality states |u + v| ≤ |u| + |v|. In geometric terms, a straight line from the start to the finish cannot be longer than a two-segment journey taking a detour.

For example, let u = 3 + 4i and v = 1 − 2i. Then u + v = 4 + 2i, so |u + v| = √20 = 2√5.

Meanwhile |u| = 5 and |v| = √5. The inequality reads 2√5 ≤ 5 + √5, which is true.

The triangle inequality is a useful check when a computed modulus seems impossibly large compared with the lengths of the component movements.

The related reverse triangle inequality, ||u| − |v|| ≤ |u − v|, gives a lower bound on distance, though a full proof may go beyond a particular question. The principle remains geometric: points cannot be closer than the difference between their distances from the same origin.

Does complex distance use a different rule from vectors?

No. In two dimensions, z = x + yi corresponds to the position vector (x, y). Its modulus is the vector’s magnitude. The difference z₁ − z₂ corresponds to the displacement vector from z₂ to z₁.

If a student has already studied displacement vectors, the complex-number distance formula should feel familiar. What changes is the notation and the additional arithmetic properties of i.

This connection is especially useful when revising several H2 topics together. A vector problem may ask for the distance between points; an Argand problem may ask for the modulus of their difference. The underlying Euclidean calculation is the same.

For vector ratio ideas in three dimensions, see the separate JC1 H2 vector section-ratio guide.

A careful method for solving locus questions

I suggest making one decision before expanding any algebra: What distance is the modulus describing?

Read |z − w| as “distance from z to the fixed point w”. If the condition fixes that distance, expect a circle. If it bounds that distance, expect a disk or its exterior. If it equates distances from two distinct fixed points, expect a perpendicular bisector.

Then translate z = x + yi and w = a + bi into Cartesian differences. Only now should you expand or square. The expansion has a purpose: it verifies the predicted geometric object.

Finally, state the centre, radius, line or region clearly. Sketching a circle without marking its centre and radius is less informative than a concise equation with the correct interpretation.

Five common mistakes and how to repair them

Mistake: writing |3 + 4i| = 7. Repair: use Pythagoras; the modulus is five.
Mistake: giving a negative modulus. Repair: a modulus is a magnitude and cannot be negative.
Mistake: forgetting brackets in z − (a + bi). Repair: subtract real and imaginary coordinates separately.
Mistake: calling |z − w| = r a filled region. Repair: equality gives only the boundary circle when r > 0; ≤ gives the filled region.
Mistake: treating equal moduli as equal complex numbers. Repair: draw two points at the same distance from the origin but in different directions.

A tutor should identify which mistake appeared first. If the problem is only a sign in subtracting complex numbers, more locus theory is unnecessary. If the learner cannot explain the meaning of a modulus, algebra drills alone will not rebuild the geometry.

Ten practice questions with answers

Question 1. Find |6 + 8i|. Answer: √(36 + 64) = 10.

Question 2. Find |−3 − 4i|. Answer: √(9 + 16) = 5.

Question 3. Find the distance between z₁ = 2 + i and z₂ = 5 + 5i. Answer: |(2 + i) − (5 + 5i)| = |−3 − 4i| = 5.

Question 4. What point is represented by z = −2 + 7i? Answer: (−2, 7) on the Argand diagram.

Question 5. Describe |z − 3| = 2. Answer: The circle centred at (3, 0) with radius two.

Question 6. Describe |z + 1 − 2i| = 4. Answer: The circle centred at (−1, 2) with radius four because the expression is z − (−1 + 2i).

Question 7. Describe |z − i| < 3. Answer: The interior of the circle centred at (0, 1), radius three; the boundary is excluded.

Question 8. Solve |z − (1 + 2i)| = 0. Answer: z = 1 + 2i only.

Question 9. What is the locus |z − (1 + i)| = |z − (5 + i)|? Answer: The line Re(z) = 3.

Question 10. If z = x + yi, what is |z − z̄|? Answer: 2|y|.

A learner who can give the answer and identify its centre, coordinate difference or geometric meaning is doing more than routine manipulation.

A three-part tuition diagnostic

First, take a small complex number such as 3 + 4i and ask for its plotted point and distance from the origin. If the learner understands the 3–4–5 right triangle, introduce a second point and ask for distance between them.

Second, ask for a circle described by |z − w| = r. Let the learner point to the centre w and label the radius r without expanding anything. Then use algebra to verify the Cartesian equation.

Third, introduce equal distances from two fixed points. A student who recognises the perpendicular bisector has begun to integrate complex notation and ordinary geometry.

The final step is independent transfer. Give an unfamiliar pair of complex numbers and ask the learner to explain the locus or distance without a teacher supplying the first equation.

Frequently asked questions

Why is the modulus always real?

Because it is √(x² + y²), the square root of a sum of non-negative real squares. It represents a non-negative Euclidean distance.

Is |z₁ − z₂| the same as |z₂ − z₁|?

Yes. The differences are negatives of each other and therefore have the same magnitude.

What does |z − w| = 5 mean?

It describes every point z exactly five units from the fixed complex point w. On an Argand diagram, this is a circle centred at w with radius five.

Does the Argand diagram require polar form?

No. Coordinates x and y and the modulus calculation are enough for these examples. The 2027 H2 Mathematics syllabus explicitly excludes complex numbers expressed in polar or exponential form, so this guide stays within Cartesian and Argand interpretations.

Can two different complex numbers have the same modulus?

Yes. Every point on a circle centred at the origin has the same modulus, so there are infinitely many possibilities for any positive radius.

Is this a JC1-only topic?

It is part of the H2 Mathematics two-year subject. Different colleges can distribute lessons differently between JC1 and JC2. The level in the title describes the tutoring audience, not a guaranteed school teaching date.

JC1 H2 Mathematics Tutor Bukit Timah | Distance Between Complex Numbers on an Argand Diagram