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Roots of Complex Numbers — Full Worked Solutions

1. Find all cube roots of 8. Fluency

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   Write 8 = 8(cos 0 + i sin 0) = 8e^i·0. So r = 8, θ = 0, n = 3.

   Formula: z_k = 8^1/3(cos(2πk/3) + i sin(2πk/3)) = 2e^2πik/3, k = 0, 1, 2.

   k = 0: z₀ = 2(cos 0 + i sin 0) = 2

   k = 1: z₁ = 2(cos(2π/3) + i sin(2π/3)) = 2(−1/2 + i√3/2) = −1 + i√3

   k = 2: z₂ = 2(cos(4π/3) + i sin(4π/3)) = 2(−1/2 − i√3/2) = −1 − i√3

   Verify: 2³ = 8 ✓. (−1+i√3)³: modulus 2, argument 2π/3, so modulus³=8 and argument 3×2π/3 = 2π ≡ 0 ✓

2. Find all 4th roots of −16. Fluency

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   Write −16 = 16(cosπ + i sinπ) = 16e^iπ. So r = 16, θ = π, n = 4.

   z_k = 16^1/4e^i(π+2πk)/4 = 2e^i(π+2πk)/4, k = 0, 1, 2, 3.

   k = 0: 2(cos(π/4) + i sin(π/4)) = 2(√2/2 + i√2/2) = √2 + i√2

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

   k = 2: 2(cos(5π/4) + i sin(5π/4)) = 2(−√2/2 − i√2/2) = −√2 − i√2

   k = 3: 2(cos(7π/4) + i sin(7π/4)) = 2(√2/2 − i√2/2) = √2 − i√2

   Check: (√2+i√2)⁴ = (2e^iπ/4)⁴ = 16e^iπ = −16 ✓

3. Find all 6th roots of unity. Fluency

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   1 = e^i·0. Roots: z_k = e^2πik/6 = e^iπk/3, k = 0, 1, 2, 3, 4, 5.

   k=0: e⁰ = 1

   k=1: e^iπ/3 = cos(π/3)+i sin(π/3) = 1/2 + i√3/2

   k=2: e^2iπ/3 = −1/2 + i√3/2

   k=3: e^iπ = −1

   k=4: e^4iπ/3 = −1/2 − i√3/2

   k=5: e^5iπ/3 = 1/2 − i√3/2

   These six points lie on the unit circle, separated by 60° = π/3 radians, forming a regular hexagon.

4. Solve z³ = 1 + i√3. Fluency

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   Convert: |1+i√3| = √(1²+(√3)²) = √4 = 2. arg = arctan(√3/1) = π/3.

   So 1+i√3 = 2e^iπ/3. The cube roots are:

   z_k = 2^1/3e^{i(π/3+2πk)/3} = 2^1/3e^{i(π/9+2πk/3)}, k = 0, 1, 2.

   z₀ = 2^1/3e^iπ/9 (argument π/9 = 20°)

   z₁ = 2^1/3e^i7π/9 (argument 7π/9 = 140°)

   z₂ = 2^1/3e^i13π/9 (argument 13π/9 = 260°)

   All have modulus 2^1/3 = ∛2 ≈ 1.260, separated by 2π/3 = 120°. They form an equilateral triangle.

5. Show that cube roots of any nonzero complex number form an equilateral triangle. Understanding

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   Let w = re^iθ with r > 0. The cube roots are z_k = r^1/3e^i(θ+2πk)/3, k = 0, 1, 2.

   Step 1 — Equal moduli: |z_k| = r^1/3 for all k. All three roots lie on the circle of radius r^1/3 centred at the origin.

   Step 2 — Equal angular spacing: arg(z₀) = θ/3, arg(z₁) = θ/3 + 2π/3, arg(z₂) = θ/3 + 4π/3. Consecutive arguments differ by exactly 2π/3 radians (120°).

   Conclusion: Three points on a circle equally spaced at 120° are exactly the vertices of an equilateral triangle inscribed in that circle. Since both conditions hold for any nonzero w, the three cube roots always form an equilateral triangle. ✓

6. Show that 1 + ω + ω² + ω³ + ω⁴ = 0 where ω = e^2πi/5. Understanding

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   Method 1 (Polynomial):

   The 5th roots of unity satisfy z⁵ = 1, i.e. z⁵ − 1 = 0.

   Factor: z⁵ − 1 = (z − 1)(z⁴ + z³ + z² + z + 1).

   Since ω = e^2πi/5 ≠ 1, it satisfies z⁴ + z³ + z² + z + 1 = 0, i.e. 1 + ω + ω² + ω³ + ω⁴ = 0. ✓

   Method 2 (Geometric series):

   S = ∑_k=0⁴ ω^k = (1 − ω⁵)/(1 − ω).

   Since ω⁵ = (e^2πi/5)⁵ = e^2πi = 1 and ω ≠ 1: S = (1−1)/(1−ω) = 0. ✓

7. Find all solutions to z⁴ + 1 = 0. Understanding

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   z⁴ = −1 = e^iπ. Apply the nth root formula with n=4, r=1, θ=π:

   z_k = 1^1/4e^i(π+2πk)/4 = e^i(2k+1)π/4, k = 0, 1, 2, 3.

   k=0: e^iπ/4 = (cosπ/4 + i sinπ/4) = (√2 + i√2)/2

   k=1: e^i3π/4 = (cos3π/4 + i sin3π/4) = (−√2 + i√2)/2

   k=2: e^i5π/4 = (−√2 − i√2)/2

   k=3: e^i7π/4 = (√2 − i√2)/2

   All four roots lie on the unit circle, forming a square rotated 45° from the axes.

   Note: z⁴+1 = (z²+√2z+1)(z²−√2z+1) over ℝ — the roots are complex conjugate pairs.

8. Find all primitive 6th roots of unity. Understanding

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   The 6th roots of unity are z_k = e^iπk/3, k = 0, 1, 2, 3, 4, 5.

   A root z_k is a primitive 6th root of unity if it has order exactly 6, i.e. z_k⁶ = 1 but z_k^m ≠ 1 for m = 1, 2, 3.

   Equivalently, this holds iff gcd(k, 6) = 1:

   • k=0: z₀=1, order 1. Not primitive.
   • k=1: gcd(1,6)=1. Primitive. z₁ = 1/2 + i√3/2
   • k=2: gcd(2,6)=2. Order 3. Not primitive.
   • k=3: gcd(3,6)=3. Order 2. Not primitive.
   • k=4: gcd(4,6)=2. Order 3. Not primitive.
   • k=5: gcd(5,6)=1. Primitive. z₅ = 1/2 − i√3/2

   There are φ(6) = 2 primitive 6th roots of unity, where φ is Euler’s totient function.

9. Find all complex z with |z|=2 and z³ a negative real. Problem Solving

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   Write z = 2e^iθ. Then z³ = 8e^3iθ.

   For z³ to be real: Im(e^3iθ) = sin(3θ) = 0, so 3θ = mπ for integer m.

   For z³ to be negative real: cos(3θ) < 0, so 3θ = (2m+1)π for integer m, i.e. 3θ = π, 3π, 5π, …

   Taking θ ∈ [0, 2π): 3θ = π, 3π, 5π ⇒ θ = π/3, π, 5π/3.

   z₀ = 2e^iπ/3 = 1 + i√3

   z₁ = 2e^iπ = −2

   z₂ = 2e^i5π/3 = 1 − i√3

   Verify z₀: z₀³ = 8e^iπ = −8 (negative real) ✓

   Verify z₁: z₁³ = 8e^3iπ = 8e^iπ = −8 (negative real) ✓

   Verify z₂: z₂³ = 8e^i5π = 8e^iπ = −8 (negative real) ✓

10. 4th roots of w: sum and product. Problem Solving

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    (a) Show sum = 0 for any nonzero w:

    Let w = re^iφ. Roots: z_k = r^1/4e^i(φ+2πk)/4, k=0,1,2,3.

    Sum = r^1/4e^iφ/4(1 + e^iπ/2 + e^iπ + e^i3π/2)

    = r^1/4e^iφ/4(1 + i + (−1) + (−i))

    = r^1/4e^iφ/4 × 0 = 0 ✓

    (b) Product when w = 1+i:

    The roots satisfy z⁴ − (1+i) = 0. By Vieta’s formulas for a monic degree-4 polynomial z⁴ + 0·z³ + 0·z² + 0·z − (1+i) = 0:

    Product of roots = (−1)⁴ × (−(1+i))/1 = −(1+i) = −1 − i

    Alternatively: product = z₀z₁z₂z₃ = (r^1/4)⁴e^{i(sum of arguments)}

    = r × e^{i(φ/4 + (φ+2π)/4 + (φ+4π)/4 + (φ+6π)/4)}

    = r × e^{i(4φ/4 + 12π/4)} = r e^i(φ+3π)

    For w = 1+i: r = √2, φ = π/4.

    Product = √2 e^i(π/4+3π) = √2 e^i13π/4

    13π/4 − 2π = 5π/4. So product = √2 e^i5π/4 = √2(cos(5π/4)+i sin(5π/4))

    = √2(−√2/2 − i√2/2) = −1 − i ✓