Complex Numbers — Topic Review — Solutions
This review covers all five lessons in the Complex Numbers topic: Cartesian Form, Operations with Complex Numbers, the Argand Diagram and Modulus, Polar Form and De Moivre’s Theorem, and Polynomial Equations with Complex Roots. Questions are exam-style and increase in difficulty. Click each question to reveal the full worked solution.
Mixed Review Questions
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Fluency
Q1 — Cartesian Form Basics
For z = −4 + 7i, state: (a) Re(z) (b) Im(z) (c) z̅ (d) z · z̅
(a) Re(z) = −4
(b) Im(z) = 7
(c) z̅ = −4 − 7i
(d) z · z̅ = (−4)² + 7² = 16 + 49 = 65
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Fluency
Q2 — Operations with Complex Numbers
Let w = 3 + 2i and z = 1 − 4i. Calculate: (a) w + z (b) wz (c) w/z (write in the form a + bi)
(a) w + z = (3+1) + (2−4)i = 4 − 2i
(b) wz = (3+2i)(1−4i) = 3 − 12i + 2i − 8i² = 3 − 10i + 8 = 11 − 10i
(c) w/z = (3+2i)/(1−4i). Multiply numerator and denominator by the conjugate of z:
= (3+2i)(1+4i) / ((1−4i)(1+4i)) = (3+12i+2i+8i²) / (1+16)
= (3+14i−8) / 17 = (−5+14i)/17 = −5/17 + (14/17)i
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Fluency
Q3 — Modulus and Argument
For z = −1 + √3 i, find: (a) |z| (b) arg(z) (in exact radians, principal value)
(a) |z| = √((−1)² + (√3)²) = √(1+3) = 2
(b) The point (−1, √3) is in the second quadrant.
Reference angle: arctan(√3/1) = π/3
Since the point is in Q2: arg(z) = π − π/3 = 2π/3
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Fluency
Q4 — Polar Form
Write z = 2(cos(π/4) + i sin(π/4)) in Cartesian form a + bi.
z = 2 cos(π/4) + 2i sin(π/4)
= 2 × (1/√2) + 2i × (1/√2)
= √2 + √2 i
z = √2 + √2 i
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Fluency
Q5 — Conjugate Root Theorem
A polynomial with real coefficients has roots 2 − 3i and −5. State all roots and write the minimal monic polynomial of degree 3.
By the conjugate root theorem, 2 + 3i is also a root.
Roots: 2+3i, 2−3i, −5.
Quadratic from complex pair (a=2, b=3): x² − 4x + (4+9) = x² − 4x + 13
P(x) = (x+5)(x²−4x+13)
= x³ − 4x² + 13x + 5x² − 20x + 65
P(x) = x³ + x² − 7x + 65
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Understanding
Q6 — De Moivre’s Theorem
Use De Moivre’s Theorem to find (1 + i)8. Express in Cartesian form.
Convert 1 + i to polar form:
|1+i| = √(1+1) = √2, arg(1+i) = π/4
1 + i = √2 cis(π/4)
Apply De Moivre’s Theorem: (r cisθ)n = rn cis(nθ)
(1+i)8 = (√2)8 cis(8 × π/4) = 24 cis(2π) = 16 × (cos 2π + i sin 2π)
= 16 × (1 + 0i) = 16
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Understanding
Q7 — Argand Diagram Geometry
On an Argand diagram, the point representing z lies on the circle |z − 2| = 3. Describe this set geometrically and determine whether z = 5 + 0i lies on this circle.
|z − 2| = 3 means the distance from z to the point 2 + 0i is 3.
Geometrically: a circle centred at (2, 0) with radius 3.
Check z = 5: |5 − 2| = |3| = 3. Yes, 3 = 3, so z = 5 lies on the circle ✓.
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Understanding
Q8 — Finding Complex Roots of a Quadratic
Find all complex roots of z² − (3+i)z + (2+i) = 0.
Try factoring: look for two numbers that multiply to 2+i and add to 3+i.
Try z = 2 and z = 1+i:
Sum: 2 + (1+i) = 3+i ✓
Product: 2(1+i) = 2+2i ≠ 2+i ×
Try z = 1 and z = 2+i:
Sum: 1 + (2+i) = 3+i ✓
Product: 1 × (2+i) = 2+i ✓
So z² − (3+i)z + (2+i) = (z − 1)(z − (2+i))
Roots: z = 1 and z = 2 + i
Note: the conjugate root theorem does NOT apply here because the coefficients are not all real.
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Understanding
Q9 — Multiplication in Polar Form
Let z1 = 4 cis(π/6) and z2 = 3 cis(π/3). Find: (a) z1z2 (b) z1/z2 Give answers in polar and Cartesian form.
(a) z1z2 = (4)(3) cis(π/6 + π/3) = 12 cis(π/6 + 2π/6) = 12 cis(π/2)
Cartesian: 12(cos(π/2) + i sin(π/2)) = 12(0 + i) = 12i
(b) z1/z2 = (4/3) cis(π/6 − π/3) = (4/3) cis(−π/6)
Cartesian: (4/3)(cos(−π/6) + i sin(−π/6)) = (4/3)(√3/2 − i/2)
= 2√3/3 − (2/3)i
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Understanding
Q10 — Cube Roots of Unity
Find all cube roots of 1 (solutions of z³ = 1) and plot them on a sketch of the Argand plane.
Write 1 = 1 cis(0). The cube roots are 11/3 cis(2kπ/3) for k = 0, 1, 2.
k = 0: 1 cis(0) = 1 (real root)
k = 1: 1 cis(2π/3) = cos(2π/3) + i sin(2π/3) = −1/2 + (√3/2)i = ω
k = 2: 1 cis(4π/3) = cos(4π/3) + i sin(4π/3) = −1/2 − (√3/2)i = ω² = ω̅
The three roots lie at equal intervals of 120° on the unit circle, forming an equilateral triangle. The two non-real roots are complex conjugates of each other.
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Understanding
Q11 — Proving a Polynomial Identity
Given that z = 1 + i, show that z4 = −4.
Method 1 (direct expansion):
z² = (1+i)² = 1 + 2i + i² = 1 + 2i − 1 = 2i
z4 = (z²)² = (2i)² = 4i² = 4(−1) = −4 ✓
Method 2 (polar form):
1+i = √2 cis(π/4)
(1+i)4 = (√2)4 cis(4 × π/4) = 4 cis(π) = 4(−1) = −4 ✓
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Problem Solving
Q12 — Solving a Quartic by Factoring
Solve z4 + 13z² + 36 = 0 over ℂ.
Let u = z². The equation becomes a quadratic in u:
u² + 13u + 36 = 0
(u + 4)(u + 9) = 0
u = −4 or u = −9
From z² = −4: z = ±√(−4) = ±2i
From z² = −9: z = ±√(−9) = ±3i
All four roots: z = 2i, −2i, 3i, −3i
z4 + 13z² + 36 = (z²+4)(z²+9)
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Problem Solving
Q13 — Locus on the Argand Plane
Find the Cartesian equation of the locus of points z = x + iy satisfying arg(z − 1) = π/4.
z − 1 = (x−1) + iy. The argument equals π/4, so:
tan(π/4) = y/(x−1) → 1 = y/(x−1) → y = x − 1
However, arg must equal π/4 (not 5π/4), so z − 1 must be in the first quadrant: x − 1 > 0 and y > 0, i.e., x > 1.
Locus: y = x − 1 with x > 1 (a ray from the point (1, 0), not including the point itself, at 45°).
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Problem Solving
Q14 — Gaussian Integer Division
Express (3 + 11i) / (1 + 2i) in the form a + bi and verify your answer by multiplication.
Multiply numerator and denominator by the conjugate of the denominator:
(3+11i)/(1+2i) × (1−2i)/(1−2i) = (3+11i)(1−2i) / (1+4)
Numerator: 3 − 6i + 11i − 22i² = 3 + 5i + 22 = 25 + 5i
= (25 + 5i) / 5 = 5 + i
Verify: (5+i)(1+2i) = 5 + 10i + i + 2i² = 5 + 11i − 2 = 3 + 11i ✓
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Problem Solving
Q15 — Fifth Roots of a Complex Number
Find all fifth roots of −32 (i.e., solve z5 = −32) and express in exact polar form cisθ.
Write −32 = 32 cis(π) (modulus 32, argument π).
The fifth roots are: z = 321/5 cis((π + 2kπ)/5) for k = 0, 1, 2, 3, 4.
321/5 = 2, since 25 = 32.
k = 0: 2 cis(π/5)
k = 1: 2 cis(3π/5)
k = 2: 2 cis(π) = −2 (the real root)
k = 3: 2 cis(7π/5) = 2 cis(−3π/5)
k = 4: 2 cis(9π/5) = 2 cis(−π/5)
The five roots are equally spaced at 72° intervals on a circle of radius 2. The complex roots come in conjugate pairs: cis(π/5) with cis(−π/5), and cis(3π/5) with cis(−3π/5).