Solutions — The Argand Diagram and Modulus

Q1 — Modulus Calculations

Use |z| = √(a² + b²) where z = a + bi.

(a) z = 3 + 4i: |z| = √(9 + 16) = √25 = 5

(b) z = −5 + 12i: |z| = √(25 + 144) = √169 = 13

(c) z = 1 − i: |z| = √(1 + 1) = √2

(d) z = −6 = −6 + 0i: |z| = √(36 + 0) = 6

Q2 — Principal Argument

Find the reference angle α = tan⁻¹(|b|/|a|) then adjust for the correct quadrant.

(a) z = 1 + i is in Quadrant 1 (Re > 0, Im > 0). tanα = 1/1 = 1 → α = π/4. arg(z) = π/4

(b) z = −√3 + i is in Quadrant 2 (Re < 0, Im > 0). tanα = 1/√3 → α = π/6. arg(z) = π − π/6 = 5π/6

(c) z = −1 − i is in Quadrant 3 (Re < 0, Im < 0). tanα = 1/1 → α = π/4. arg(z) = α − π = −3π/4

(d) z = 2i lies on the positive imaginary axis. arg(z) = π/2

Q3 — Modulus Properties

Apply the modulus multiplication/division rules.

(a) |zw| = |z| × |w| = 3 × 4 = 12

(b) |z/w| = |z|/|w| = 3/4 = 3/4

(c) |z²| = |z|² = 3² = 9

(d) By the triangle inequality: |z + w| ≤ |z| + |w| = 3 + 4 = 7. The maximum 7 is achieved when z and w point in the same direction on the Argand diagram.

Q4 — Plotting on the Argand Diagram

z = 2 + 3i → point (2, 3)

z̅ = 2 − 3i → point (2, −3): reflection of z across the real axis

−z = −2 − 3i → point (−2, −3): rotation of z by 180° about the origin (reflection through the origin)

−z̅ = −2 + 3i → point (−2, 3): reflection of z across the imaginary axis

All four points have the same modulus |z| = √(4 + 9) = √13, and they form the four corners of a rectangle with centre at the origin.

Q5 — Modulus and Argument Together

z = −2 + 2i. Here a = −2, b = 2.

Modulus: |z| = √(4 + 4) = √8 = 2√2

Argument: z is in Quadrant 2 (Re < 0, Im > 0). Reference angle: tanα = 2/2 = 1 → α = π/4. So arg(z) = π − π/4 = 3π/4.

Verification: 2√2 (cos(3π/4) + i sin(3π/4)) = 2√2 × (−1/√2) + 2√2 × (1/√2) i = −2 + 2i ✓

Q6 — Distance Between Two Complex Numbers

The distance between z and w in the complex plane equals |z − w|.

z − w = (3 + i) − (−1 + 4i) = (3 − (−1)) + (1 − 4)i = 4 − 3i

|z − w| = √(4² + 3²) = √(16 + 9) = √25 = 5

Geometrically: z = (3, 1) and w = (−1, 4) in the plane. The horizontal separation is 4, vertical separation is 3 — a classic 3-4-5 right triangle.

Q7 — Geometric Locus

(a) |z| = 5 means the distance from z to the origin equals 5. This describes a circle centred at the origin with radius 5. For z = a + bi: a² + b² = 25.

(b) |z − 2i| represents the distance from z to the point 2i (i.e., the point (0, 2) on the Argand diagram). Setting this equal to 3 gives a circle centred at (0, 2) with radius 3. For z = a + bi: a² + (b − 2)² = 9.

Q8 — Argument of a Product

When two complex numbers are multiplied, their arguments add (this is made precise in polar form).

arg(zw) = arg(z) + arg(w) = π/3 + π/4 = 4π/12 + 3π/12 = 7π/12

Geometric meaning: Multiplying z by w rotates z anticlockwise by arg(w) = π/4. Multiplication by a complex number is a combined rotation and scaling: the angle of zw equals the sum of the angles, and the modulus of zw equals the product of the moduli. This is the central reason complex numbers are so powerful for describing rotations.

Q9 — Finding Complex Numbers from Geometric Conditions

Let z = a + bi. We need |z| = 5 and Re(z) = 3, so a = 3.

|z| = √(3² + b²) = 5

9 + b² = 25

b² = 16 → b = ±4

Solutions: z = 3 + 4i and z = 3 − 4i

Geometric interpretation: The circle |z| = 5 (centred at origin, radius 5) intersects the vertical line Re(z) = 3 at exactly two points: (3, 4) and (3, −4). These are complex conjugates of each other — a pattern that will recur throughout the course.

Q10 — Triangle Inequality Application

The key strategy is to rewrite z + w in terms of the bounded quantities (z − 3) and (w + i).

Write: z + w = (z − 3) + (w + i) + (3 − i)

Apply the triangle inequality repeatedly:

|z + w| = |(z − 3) + (w + i) + (3 − i)|

≤ |z − 3| + |w + i| + |3 − i|

≤ 1 + 2 + √(9 + 1)

= 3 + √10

The maximum value of |z + w| is 3 + √10 (achieved when all three vectors point in the same direction in the complex plane).