Term 1 review: binomial expansions, complex numbers, and vectors, lines and planes
Week 15 · Term 1 review · 3 lessons
2
Review: complex numbers, trigonometric and exponential forms, roots and the cube roots of unity
Pure Mathematics Sec 3 Term 1, Units 1–4, Review
GoalRevise complex numbers: modulus, amplitude, trigonometric and exponential forms, powers, roots and the cube roots of unity.
New words
principal amplitude · the amplitude θ of z lying in ]−π, π]De Moivre’s theorem · (cos θ + i sin θ)ⁿ = cos nθ + i sin nθcube roots of unity · 1, ω, ω², with 1 + ω + ω² = 0 and ω³ = 1
Explanation
Weeks 4 to 7 build on one picture: a complex number x + yi is the point (x, y), with distance |z| = r and angle θ. First find r, then the reference angle α and the quadrant to get the principal amplitude, and write z = r(cos θ + i sin θ), or z = re^(iθ). In this form multiplication multiplies the moduli and adds the amplitudes, division divides and subtracts, and the power zⁿ = rⁿ(cos nθ + i sin nθ) is De Moivre’s theorem. Every amplitude must be brought back into ]−π, π] by adding or subtracting 2π. Roots: Zⁿ = a has n roots, with modulus r^(1/n) and amplitudes (θ + 360°k)/n; on the Argand plane they are the corners of a regular n-sided polygon. A square root in the form x + yi is found from x² − y² = a, 2xy = b and x² + y² = |a + bi|. The cube roots of 1 are 1, ω, ω², and ω lets you simplify long expressions: reduce powers by ω³ = 1 and replace 1 + ω + ω² by 0. When you finish, check one value in the original equation.
Let z₁ = 1 − i and z₂ = −2 + 2√3 i. |z₁| = √2 with θ₁ = −π/4, and |z₂| = 4 with θ₂ = 2π/3. Then z₁z₂ = 4√2 e^(i(−π/4 + 2π/3)) = 4√2 e^(i5π/12), and z₂/z₁ = (4/√2) e^(i(2π/3 + π/4)) = 2√2 e^(i11π/12), which is inside ]−π, π]. A check: the real part of z₁z₂ is 2√3 − 2 ≈ 1.46.
Power: (−1 + i)⁸. The modulus is √2 and the amplitude is 3π/4, so the power has modulus (√2)⁸ = 16 and amplitude 8 × 3π/4 = 6π, which is 0 after removing full turns. So (−1 + i)⁸ = 16.
Roots: Z⁴ = −16. Here −16 = 16(cos 180° + i sin 180°), so Z = 2(cos((180° + 360°k)/4) + i sin(…)) for k = 0, 1, 2, 3, with amplitudes 45°, 135°, 225°, 315°. The roots are √2 + √2 i, −√2 + √2 i, −√2 − √2 i and √2 − √2 i, the corners of a square on the circle of radius 2.
Rotate with multiplication. Plot z = 2 + i. Multiply it by i and plot the result, then by i again. What shape do the four points z, iz, i²z, i³z make, and why? (A square: each multiplication turns the point through a quarter turn.)
Practice
1
Write −3√3 − 3i in trigonometric form.
|z| = √(27 + 9) = 6. The reference angle is tan⁻¹(3/(3√3)) = π/6, and the point is in the third quadrant, so θ = π/6 − π = −5π/6. z = 6(cos(−5π/6) + i sin(−5π/6)).
2
Let z = 2(cos 40° + i sin 40°) and w = 3(cos 80° + i sin 80°). Find zw and z/w.
zw = 6(cos 120° + i sin 120°) = −3 + 3√3 i. z/w = (2/3)(cos(−40°) + i sin(−40°)).
3
Find (1 + √3 i)⁶.
|1 + √3 i| = 2 and θ = π/3, so the power is 2⁶(cos 2π + i sin 2π) = 64.
4
Solve Z³ = 8i and give the roots in the form x + yi.
8i = 8(cos 90° + i sin 90°), so Z = 2(cos((90° + 360°k)/3) + i sin(…)): amplitudes 30°, 150°, 270°. The roots are √3 + i, −√3 + i and −2i.
5
Simplify (1 + ω²)³ + (1 + ω)³.
1 + ω² = −ω and 1 + ω = −ω², so the sum is (−ω)³ + (−ω²)³ = −1 − 1 = −2.