MathsCBSE Class 11 · Complex Numbers and Quadratic Equations

Modulus, Conjugate and the Argand Plane: Properties and Solved Questions

Updated 17 October 2026 · Fact-checked

For z = a + ib, the conjugate is a − ib and the modulus is |z| = √(a² + b²), the distance of the point (a, b) from the origin. Key results: z·z̄ = |z|², |z1z2| = |z1||z2|, |z1/z2| = |z1|/|z2| and |z1 + z2| ≤ |z1| + |z2|. On the Argand plane, |z − w| is the distance between the points z and w, so |z − w| = r describes a circle.

Understand Modulus, Conjugate and the Argand Plane

Every complex number $z = x + i y$ corresponds to the point $(x, y)$ of the Argand plane. Algebra and geometry then translate into each other.

Conjugate. $\overset{z}{ˉ} = x - i y$ is the mirror image of $z$ in the real axis. Facts: $\overset{ˉ}{\overset{z}{ˉ}} = z$ , $\overline{z_{1} \pm z_{2}} = \overset{z}{ˉ}_{1} \pm \overset{z}{ˉ}_{2}$ , $\overline{z_{1} z_{2}} = \overset{z}{ˉ}_{1} \overset{z}{ˉ}_{2}$ , $\overline{(z_{1} / z_{2})} = \overset{z}{ˉ}_{1} / \overset{z}{ˉ}_{2}$ ; $z + \overset{z}{ˉ} = 2 Re z$ , $z - \overset{z}{ˉ} = 2 i Im z$ . A number is real iff $z = \overset{z}{ˉ}$ and purely imaginary iff $z = - \overset{z}{ˉ}$ (and $z \neq = 0$ ).

Modulus. $∣ z ∣ = x^{2} + y^{2} = z \overset{z}{ˉ} \geq 0$ , the distance of the point from the origin. Facts:

$∣ z_{1} z_{2} ∣ = ∣ z_{1} ∣∣ z_{2} ∣$ and $\frac{z _{1}}{z _{2}} = \frac{∣ z _{1} ∣}{∣ z _{2} ∣}$ (so the multiplicative property carries over to products of many factors, and $∣ z^{n} ∣ = ∣ z ∣^{n}$ ).
$∣ \overset{z}{ˉ} ∣ = ∣ z ∣ = ∣ - z ∣$ .
Triangle inequality: $∣ z_{1} + z_{2} ∣ \leq ∣ z_{1} ∣ + ∣ z_{2} ∣$ and $∣ z_{1} ∣ - ∣ z_{2} ∣ \leq ∣ z_{1} - z_{2} ∣$ .
Parallelogram law: $∣ z_{1} + z_{2} ∣^{2} + ∣ z_{1} - z_{2} ∣^{2} = 2 (∣ z_{1} ∣^{2} + ∣ z_{2} ∣^{2})$ .

Distance and loci. $∣ z_{1} - z_{2} ∣$ is the distance between the points $z_{1}$ and $z_{2}$ . Hence:

$∣ z - z_{0} ∣ = r$ : circle with centre $z_{0}$ and radius $r$ ;
$∣ z - z_{1} ∣ = ∣ z - z_{2} ∣$ : the perpendicular bisector of the segment joining $z_{1}, z_{2}$ ;
$∣ z ∣ = R$ and a fixed point $w$ : the distance $∣ z - w ∣$ varies between $∣ w ∣ - R$ and $∣ w ∣ + R$ .

To find a locus algebraically, put $z = x + i y$ , square both sides and simplify.

Multiplying by i. $i z$ turns the point $z$ through $9 0^{\circ}$ anticlockwise about the origin, keeping $∣ i z ∣ = ∣ z ∣$ .

Polar form is taught only formatively in 2026-27, so this page works with $x + i y$ throughout.

Modulus, Conjugate and the Argand Plane: formulas to remember

Modulus: $∣ z ∣ = \sqrt (x^{2} + y^{2})$; Distance from the origin.
Conjugate product: $z \times \overset{z}{ˉ} = ∣ z ∣^{2}$; Real and non-negative.
Product and quotient: $∣ z 1 z 2∣ = ∣ z 1∣∣ z 2∣; ∣ z 1/ z 2∣ = ∣ z 1∣/∣ z 2∣$; For z2 not equal to 0.
Triangle inequality: $∣ z 1 + z 2∣ \leq ∣ z 1∣ + ∣ z 2∣$; Also ||z1| − |z2|| ≤ |z1 − z2|.
Parallelogram law: $∣ z 1 + z 2 ∣^{2} + ∣ z 1 - z 2 ∣^{2} = 2 (∣ z 1 ∣^{2} + ∣ z 2 ∣^{2})$; Expand using z z̄.
Circle: $∣ z - z 0∣ = r$; Centre z0, radius r.

How to solve Modulus, Conjugate and the Argand Plane questions

Use these steps for modulus calculations, proofs and loci.

1Decide whether to compute $∣ z ∣$ directly from $x + i y$ or to use $∣ z_{1} z_{2} ∣ = ∣ z_{1} ∣∣ z_{2} ∣$ (usually quicker for fractions and powers).
2For proofs, replace $∣ z ∣^{2}$ by $z \overset{z}{ˉ}$ and use the conjugate rules.
3For a locus, write $z = x + i y$ , substitute into the given condition, square both sides to remove roots and simplify to an equation in $x$ and $y$ .
4Name the curve (line, circle) and give its centre or equation.
5For maximum and minimum of $∣ z - w ∣$ with $∣ z ∣ = R$ , use the triangle inequality or the geometry of the circle: the extreme values are $∣ w ∣ \pm R$ (with $∣ w ∣ > R$ ).
6Test one point to check the final locus.

Modulus, Conjugate and the Argand Plane in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Modulus as straight-line distance

Suppose a treasure map uses the complex plane with 1 unit = 1 km. The starting hut is at $0$ and the treasure at $3 + 4 i$ .

How far is the treasure from the hut, and where is its reflection in the east-west axis (the real axis)?

Try it first, then show the solutionSolution

Distance $= ∣3 + 4 i ∣ = 9 + 16 = 5$ km.
Reflection in the real axis is the conjugate: $3 - 4 i$ , 4 km south of the east-west line.
Distance of the reflected point from the hut is also $∣3 - 4 i ∣ = 5$ km.

Answer: 5 km; the reflection is $3 - 4 i$ .

Memory hook: Modulus is distance; conjugate is a mirror.

Locus of a circle of coverage

Suppose a radio mast stands at the point $1 - 2 i$ on a map and its signal reaches exactly 3 km.

Write the boundary of the covered region as a condition on $z = x + i y$ and find its equation.

Try it first, then show the solutionSolution

A point $z$ is exactly 3 km from the mast when $∣ z - (1 - 2 i) ∣ = 3$ .
$∣ (x - 1) + i (y + 2) ∣ = 3$ gives $(x - 1)^{2} + (y + 2)^{2} = 9$ .
This is a circle of centre $(1, - 2)$ and radius 3.

Answer: $(x - 1)^{2} + (y + 2)^{2} = 9$

Memory hook: |z − a| = r: all points at distance r from a.

Tricks to remember Modulus, Conjugate and the Argand Plane

Short memory aids. Say each one aloud once, then use it on a question the same day.

Modulus of a fraction
Do not simplify first: |numerator| ÷ |denominator|.
Use it when: Fractions and powers.
z z-bar is the squared length
Replace |z|² by z × z̄ when proving identities.
Use it when: Proof questions.
Difference means distance
|z − w| = the distance between the points z and w.
Use it when: Loci and extreme values.
Extremes on a circle
For |z| = R and fixed w, the extremes of |z − w| are |w| + R and ||w| − R|.
Use it when: Maximum and minimum questions.

Modulus, Conjugate and the Argand Plane: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Modulus and conjugate of a quotient · 2 marks · Very short answerBoard level

Find the modulus and the conjugate of $\frac{3 + 4 i}{1 - 2 i}$ .

Try it first, then show the solutionSolution

$\frac{3 + 4 i}{1 - 2 i} = \frac{( 3 + 4 i ) ( 1 + 2 i )}{5} = \frac{3 + 6 i + 4 i - 8}{5} = - 1 + 2 i$ . (1 mark)
Modulus $= 1 + 4 = 5$ (check: $\frac{5}{5} = 5$ ); conjugate $= - 1 - 2 i$ . (1 mark)

Answer: Modulus $5$ ; conjugate $- 1 - 2 i$

Example 2: Product and quotient moduli · 3 marks · Short answerBoard level

For $z_{1} = 3 + 4 i$ and $z_{2} = 5 - 12 i$ find $∣ z_{1} z_{2} ∣$ and $\frac{z _{1}}{z _{2}}$ , and verify the first by multiplying out.

Try it first, then show the solutionSolution

$∣ z_{1} ∣ = 5$ , $∣ z_{2} ∣ = 25 + 144 = 13$ . (½ mark)
$∣ z_{1} z_{2} ∣ = 65$ and $\frac{z _{1}}{z _{2}} = \frac{5}{13}$ . (1 mark)
Direct: $z_{1} z_{2} = 15 - 36 i + 20 i - 48 i^{2} = 63 - 16 i$ . (1 mark)
$∣63 - 16 i ∣ = 3969 + 256 = 4225 = 65$ . Verified. (½ mark)

Answer: 65 and $\frac{5}{13}$

Example 3: A locus · 3 marks · Short answerHarder board level

Find the locus of $z$ if $∣ z - 3∣ = ∣ z + 3 i ∣$ .

Try it first, then show the solutionSolution

Put $z = x + i y$ : $∣ (x - 3) + i y ∣ = ∣ x + i (y + 3) ∣$ . (½ mark)
Square: $(x - 3)^{2} + y^{2} = x^{2} + (y + 3)^{2}$ . (1 mark)
$- 6 x + 9 = 6 y + 9$ , so $y = - x$ , i.e. $x + y = 0$ . (1 mark)
This is the perpendicular bisector of the segment joining $3$ and $- 3 i$ ; it is the line $y = - x$ through the origin. (½ mark)

Answer: The line $x + y = 0$

Example 4: Extreme distances · 3 marks · Short answerHarder board level

If $∣ z ∣ = 2$ , find the greatest and least values of $∣ z - 3 + 4 i ∣$ .

Try it first, then show the solutionSolution

$∣ z - 3 + 4 i ∣ = ∣ z - w ∣$ with $w = 3 - 4 i$ , $∣ w ∣ = 5$ . (1 mark)
Triangle inequality: $∣ w ∣ - ∣ z ∣ \leq ∣ z - w ∣ \leq ∣ w ∣ + ∣ z ∣$ . (1 mark)
So $3 \leq ∣ z - w ∣ \leq 7$ ; both values occur at $z = \pm \frac{2}{5} (3 - 4 i)$ (on the line through $0$ and $w$ ). (1 mark)

Answer: Least 3, greatest 7

Example 5: Case study: a robot on the plane · 4 marks · Case-basedHOTS / competency

Suppose a robot starts at the origin of the Argand plane (1 unit = 1 m). It moves by $z_{1} = 3 + 4 i$ and then by $z_{2} = - 1 + 2 i$ .
(i) Find its final position. (1 mark)
(ii) Find its distance from the start. (1 mark)
(iii) A turntable rotates the first move $z_{1}$ by $9 0^{\circ}$ anticlockwise (multiplication by $i$ ). Find the new vector. (1 mark)
(iv) Show that the rotation does not change the length of $z_{1}$ . (1 mark)

Try it first, then show the solutionSolution

(i) $z_{1} + z_{2} = 2 + 6 i$ . (1 mark)
(ii) $∣2 + 6 i ∣ = 4 + 36 = 40 = 210 \approx 6.32$ m. (1 mark)
(iii) $i z_{1} = i (3 + 4 i) = - 4 + 3 i$ . (1 mark)
(iv) $∣ i z_{1} ∣ = 16 + 9 = 5 = ∣ z_{1} ∣$ , since $∣ i ∣ = 1$ . (1 mark)

Answer: (i) $2 + 6 i$ (ii) $210$ m (iii) $- 4 + 3 i$ (iv) both are 5

Where students lose marks in Modulus, Conjugate and the Argand Plane

The slips examiners see most often, and what to write instead.

Mistake: Writing |z|² = z²
z² is complex and generally not real; |z|² = z z̄.
Wrong working
$∣3 + 4 i ∣^{2} = (3 + 4 i)^{2} = - 7 + 24 i$ .
Right working
$∣3 + 4 i ∣^{2} = (3 + 4 i) (3 - 4 i) = 25$ .
Do this: Never drop the bar.
Usually costs: Method and answer.
Mistake: Adding moduli to get the modulus of a sum
|z1 + z2| is at most |z1| + |z2|, with equality only in special cases.
Wrong working
$∣3 + 4 i + 1 + 0 i ∣ = 5 + 1 = 6$ .
Right working
$∣4 + 4 i ∣ = 42 \approx 5.66$ .
Do this: Add first, then take the modulus.
Usually costs: The numerical answer.
Mistake: Forgetting to square when finding a locus
Students equate the root expressions and leave square roots in.
Wrong working
Stopping at $(x - 3)^{2} + y^{2} = x^{2} + (y + 3)^{2}$ .
Right working
Square both sides: $(x - 3)^{2} + y^{2} = x^{2} + (y + 3)^{2}$ and simplify.
Do this: Both sides are non-negative so squaring is safe.
Usually costs: 1 to 2 marks.
Mistake: Not naming the locus
The equation is found but not interpreted.
Wrong working
Ends with " $x + y = 0$ ".
Right working
"The locus is the line $x + y = 0$ , the perpendicular bisector of the join of 3 and $- 3 i$ ."
Do this: Finish with a sentence naming the curve.
Usually costs: The last mark.

Exam tips

Use $∣ z_{1} z_{2} ∣ = ∣ z_{1} ∣∣ z_{2} ∣$ to avoid multiplying out long products.
In proofs replace $∣ z ∣^{2}$ by $z \overset{z}{ˉ}$ ; this is the standard board method.
Draw a quick Argand diagram for locus and extreme-value questions; it makes the answer visible.
Polar form is formative only in 2026-27; do not spend time converting unless a question asks.
Check a locus by testing one point on it.

Practice questions on Modulus, Conjugate and the Argand Plane

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The modulus of $\frac{1 + i}{1 - i}$ is:
(a) 0
(b) 1
(c) $2$
(d) 2
Hint
$\frac{1 + i}{1 - i} = i$ .
Answer
(b). $∣ i ∣ = 1$ (or $2 / 2 = 1$ ).
Q2 · 1 mark · MCQHarder board level
If $∣ z ∣ = 1$ then $z + \frac{1}{z}$ is:
(a) purely imaginary
(b) a real number between $- 2$ and $2$ (inclusive)
(c) zero only
(d) a non-real number
Hint
For $∣ z ∣ = 1$ , $z \overset{z}{ˉ} = 1$ so $1/ z = \overset{z}{ˉ}$ .
Answer
(b). $z + \overset{z}{ˉ} = 2 Re z$ , and $- 1 \leq Re z \leq 1$ .
Q3 · 3 marks · Short answerHarder board level
Prove that $∣ z_{1} + z_{2} ∣^{2} + ∣ z_{1} - z_{2} ∣^{2} = 2 (∣ z_{1} ∣^{2} + ∣ z_{2} ∣^{2})$ and verify it for $z_{1} = 1 + i$ , $z_{2} = 2 - i$ .
Hint
Expand $(z_{1} + z_{2}) (\overset{z}{ˉ}_{1} + \overset{z}{ˉ}_{2})$ .
Answer
$∣ z_{1} \pm z_{2} ∣^{2} = ∣ z_{1} ∣^{2} + ∣ z_{2} ∣^{2} \pm (z_{1} \overset{z}{ˉ}_{2} + \overset{z}{ˉ}_{1} z_{2})$ ; adding, the cross terms cancel. Check: $z_{1} + z_{2} = 3$ , $∣ \cdot ∣^{2} = 9$ ; $z_{1} - z_{2} = - 1 + 2 i$ , $∣ \cdot ∣^{2} = 5$ ; sum $14$ . RHS $2 (2 + 5) = 14$ .
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): For every non-zero complex number $z$ , $\frac{z}{z ˉ} = 1$ .
Reason (R): $∣ \overset{z}{ˉ} ∣ = ∣ z ∣$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Use $∣ z_{1} / z_{2} ∣ = ∣ z_{1} ∣/∣ z_{2} ∣$ .
Answer
(a). $∣ z / \overset{z}{ˉ} ∣ = ∣ z ∣/∣ \overset{z}{ˉ} ∣ = ∣ z ∣/∣ z ∣ = 1$ .
Q5 · 4 marks · Short answerHOTS / competency
If $(a + ib) (c + i d) (e + i f) = A + i B$ , prove that $(a^{2} + b^{2}) (c^{2} + d^{2}) (e^{2} + f^{2}) = A^{2} + B^{2}$ , and check it for $(1 + i) (2 + i) (1 + 2 i)$ .
Hint
Take the modulus of both sides and square.
Answer
$∣ a + ib ∣∣ c + i d ∣∣ e + i f ∣ = ∣ A + i B ∣$ ; squaring gives the result. Check: $(1 + i) (2 + i) = 1 + 3 i$ ; $(1 + 3 i) (1 + 2 i) = - 5 + 5 i$ , so $A^{2} + B^{2} = 50$ ; and $(1 + 1) (4 + 1) (1 + 4) = 2 \cdot 5 \cdot 5 = 50$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What is |x + iy|?Show answer
√(x² + y²).
2.What does z z̄ equal?Show answer
|z|².
3.What is |z1 z2|?Show answer
|z1| |z2|.
4.What does |z − w| mean geometrically?Show answer
The distance between the points z and w.
5.What shape is |z − a| = r?Show answer
A circle with centre a and radius r.

Modulus, Conjugate and the Argand Plane: frequently asked questions

What is the geometric meaning of the conjugate?

It is the reflection of the point in the real axis (the x-axis).

Why is the modulus always a real number?

Because it is a square root of x² + y², which is a non-negative real number, so it is a distance.

Can the modulus of a sum exceed the sum of moduli?

No. The triangle inequality says |z1 + z2| ≤ |z1| + |z2|, because one side of a triangle cannot exceed the sum of the other two.

Is polar form needed here?

Polar form is taught only formatively in 2026-27, so the page stays with x + iy; the Argand plane and modulus are the examined ideas.