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Induction for Matrices and Complex Numbers — Full Worked Solutions

1. Verify the base case n = 1 for [[1,2],[0,1]]ⁿ = [[1,2n],[0,1]]. Fluency

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   For n = 1, the formula [[1,2n],[0,1]] gives [[1,2(1)],[0,1]] = [[1,2],[0,1]].

   Direct computation: [[1,2],[0,1]]¹ = [[1,2],[0,1]].

   LHS = RHS. The base case n = 1 is verified. ✓

2. Compute [[1,1],[0,1]]² and [[1,1],[0,1]]³ and verify the pattern. Fluency

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   Computing A² = [[1,1],[0,1]] × [[1,1],[0,1]]:

   Entry (1,1): 1(1) + 1(0) = 1

   Entry (1,2): 1(1) + 1(1) = 2

   Entry (2,1): 0(1) + 1(0) = 0

   Entry (2,2): 0(1) + 1(1) = 1

   A² = [[1,2],[0,1]]. This matches [[1,n],[0,1]] with n = 2. ✓

   Computing A³ = A² × A = [[1,2],[0,1]] × [[1,1],[0,1]]:

   Entry (1,1): 1(1) + 2(0) = 1

   Entry (1,2): 1(1) + 2(1) = 3

   Entry (2,1): 0(1) + 1(0) = 0

   Entry (2,2): 0(1) + 1(1) = 1

   A³ = [[1,3],[0,1]]. This matches [[1,n],[0,1]] with n = 3. ✓

   The pattern is confirmed: the top-right entry equals the power n.

3. Evaluate (cos30° + i sin30°)⁴ exactly using De Moivre’s theorem. Fluency

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   By De Moivre’s theorem:

   (cos30° + i sin30°)⁴ = cos(4 × 30°) + i sin(4 × 30°) = cos120° + i sin120°

   Exact values: cos120° = −1/2, sin120° = √3/2.

   Answer: −1/2 + (√3/2)i

4. Evaluate (cos(π/6) + i sin(π/6))⁶ using De Moivre’s theorem. Fluency

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   By De Moivre’s theorem:

   (cos(π/6) + i sin(π/6))⁶ = cos(6 × π/6) + i sin(6 × π/6) = cos(π) + i sin(π)

   cos(π) = −1, sin(π) = 0.

   Answer: −1

   Note: this result shows that (cos(π/6) + i sin(π/6)) is a primitive 12th root of unity, and raising it to the 6th power gives −1, which is the primitive square root of unity.

5. Prove [[1,1],[0,1]]ⁿ = [[1,n],[0,1]] for all positive integers n. Understanding

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   Claim: P(n): [[1,1],[0,1]]ⁿ = [[1,n],[0,1]] for all positive integers n.

   Base case n = 1: [[1,1],[0,1]]¹ = [[1,1],[0,1]] = [[1,1],[0,1]] (formula with n=1). LHS = RHS. ✓

   Inductive hypothesis: Assume [[1,1],[0,1]]^k = [[1,k],[0,1]] for some integer k ≥ 1.

   Inductive step: We must show [[1,1],[0,1]]^k+1 = [[1,k+1],[0,1]].

   [[1,1],[0,1]]^k+1 = [[1,1],[0,1]]^k × [[1,1],[0,1]]

   = [[1,k],[0,1]] × [[1,1],[0,1]]     [by the inductive hypothesis]

   Entry-by-entry computation:

   (1,1): 1(1) + k(0) = 1

   (1,2): 1(1) + k(1) = k + 1

   (2,1): 0(1) + 1(0) = 0

   (2,2): 0(1) + 1(1) = 1

   Therefore [[1,1],[0,1]]^k+1 = [[1, k+1],[0, 1]]. ✓

   Conclusion: By the principle of mathematical induction, [[1,1],[0,1]]ⁿ = [[1,n],[0,1]] for all positive integers n.

6. Prove [[2,0],[0,3]]ⁿ = [[2ⁿ,0],[0,3ⁿ]] for all positive integers n. Understanding

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   Claim: P(n): [[2,0],[0,3]]ⁿ = [[2ⁿ,0],[0,3ⁿ]] for all positive integers n.

   Base case n = 1: [[2,0],[0,3]]¹ = [[2,0],[0,3]] and [[2¹,0],[0,3¹]] = [[2,0],[0,3]]. ✓

   Inductive hypothesis: Assume [[2,0],[0,3]]^k = [[2^k,0],[0,3^k]] for some integer k ≥ 1.

   Inductive step: We must show [[2,0],[0,3]]^k+1 = [[2^k+1,0],[0,3^k+1]].

   [[2,0],[0,3]]^k+1 = [[2,0],[0,3]]^k × [[2,0],[0,3]]

   = [[2^k,0],[0,3^k]] × [[2,0],[0,3]]     [by the inductive hypothesis]

   Entry-by-entry computation:

   (1,1): 2^k(2) + 0(0) = 2^k+1

   (1,2): 2^k(0) + 0(3) = 0

   (2,1): 0(2) + 3^k(0) = 0

   (2,2): 0(0) + 3^k(3) = 3^k+1

   Therefore [[2,0],[0,3]]^k+1 = [[2^k+1,0],[0,3^k+1]]. ✓

   Conclusion: By the principle of mathematical induction, [[2,0],[0,3]]ⁿ = [[2ⁿ,0],[0,3ⁿ]] for all positive integers n.

7. Prove De Moivre’s theorem for positive integers. Understanding

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   Claim: P(n): (cosθ + i sinθ)ⁿ = cos(nθ) + i sin(nθ) for all positive integers n.

   Base case n = 1: (cosθ + i sinθ)¹ = cosθ + i sinθ = cos(1·θ) + i sin(1·θ). ✓

   Inductive hypothesis: Assume (cosθ + i sinθ)^k = cos(kθ) + i sin(kθ) for some integer k ≥ 1.

   Inductive step: We must show (cosθ + i sinθ)^k+1 = cos((k+1)θ) + i sin((k+1)θ).

   (cosθ + i sinθ)^k+1 = (cosθ + i sinθ)^k × (cosθ + i sinθ)

   = [cos(kθ) + i sin(kθ)](cosθ + i sinθ)     [by the inductive hypothesis]

   Expanding and using i² = −1:

   = cos(kθ)cosθ − sin(kθ)sinθ + i[sin(kθ)cosθ + cos(kθ)sinθ]

   Applying compound angle formulas:

   Real part: cos(kθ)cosθ − sin(kθ)sinθ = cos(kθ + θ) = cos((k+1)θ)

   Imaginary part: sin(kθ)cosθ + cos(kθ)sinθ = sin(kθ + θ) = sin((k+1)θ)

   Therefore (cosθ + i sinθ)^k+1 = cos((k+1)θ) + i sin((k+1)θ). ✓

   Conclusion: By the principle of mathematical induction, De Moivre’s theorem holds for all positive integers n.

8. Find the exact value of (1 + i)⁸ using De Moivre’s theorem. Understanding

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   Step 1: Convert to polar form.

   |1 + i| = √(1² + 1²) = √2

   arg(1 + i) = arctan(1/1) = π/4 (first quadrant)

   So 1 + i = √2 cis(π/4).

   Step 2: Apply De Moivre’s theorem.

   (1 + i)⁸ = [√2 cis(π/4)]⁸ = (√2)⁸ cis(8 × π/4) = 2⁴ cis(2π)

   Step 3: Convert back to Cartesian form.

   cis(2π) = cos(2π) + i sin(2π) = 1 + 0i = 1

   (1 + i)⁸ = 16 × 1 = 16

9. Prove [[1,0],[1,1]]ⁿ = [[1,0],[n,1]] and hence find [[1,0],[1,1]]¹⁰⁰. Problem Solving

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   Claim: P(n): [[1,0],[1,1]]ⁿ = [[1,0],[n,1]] for all positive integers n.

   Base case n = 1: [[1,0],[1,1]]¹ = [[1,0],[1,1]] and [[1,0],[1,1]] with n=1 gives [[1,0],[1,1]]. ✓

   Inductive hypothesis: Assume [[1,0],[1,1]]^k = [[1,0],[k,1]] for some integer k ≥ 1.

   Inductive step: We must show [[1,0],[1,1]]^k+1 = [[1,0],[k+1,1]].

   [[1,0],[1,1]]^k+1 = [[1,0],[1,1]]^k × [[1,0],[1,1]]

   = [[1,0],[k,1]] × [[1,0],[1,1]]     [by the inductive hypothesis]

   Entry-by-entry computation:

   (1,1): 1(1) + 0(1) = 1

   (1,2): 1(0) + 0(1) = 0

   (2,1): k(1) + 1(1) = k+1

   (2,2): k(0) + 1(1) = 1

   Therefore [[1,0],[1,1]]^k+1 = [[1,0],[k+1,1]]. ✓

   Conclusion: By the principle of mathematical induction, [[1,0],[1,1]]ⁿ = [[1,0],[n,1]] for all positive integers n.

   Hence: [[1,0],[1,1]]¹⁰⁰ = [[1,0],[100,1]].

10. Use De Moivre’s theorem (n=3) to derive cos(3θ) = 4cos³θ − 3cosθ. Problem Solving

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    By De Moivre’s theorem with n = 3:

    (cosθ + i sinθ)³ = cos(3θ) + i sin(3θ)     … (*).

    Expand the left side by the binomial theorem with a = cosθ, b = i sinθ:

    (cosθ + i sinθ)³

    = cos³θ + 3cos²θ(i sinθ) + 3cosθ(i sinθ)² + (i sinθ)³

    = cos³θ + 3i cos²θ sinθ + 3cosθ · i² sin²θ + i³ sin³θ

    Since i² = −1 and i³ = −i:

    = cos³θ + 3i cos²θ sinθ − 3cosθ sin²θ − i sin³θ

    = (cos³θ − 3cosθ sin²θ) + i(3cos²θ sinθ − sin³θ)

    Comparing with (*), equating real parts:

    cos(3θ) = cos³θ − 3cosθ sin²θ

    Substitute sin²θ = 1 − cos²θ:

    = cos³θ − 3cosθ(1 − cos²θ) = cos³θ − 3cosθ + 3cos³θ

    cos(3θ) = 4cos³θ − 3cosθ ✓

    Equating imaginary parts:

    sin(3θ) = 3cos²θ sinθ − sin³θ

    Substituting cos²θ = 1 − sin²θ:

    sin(3θ) = 3sinθ − 4sin³θ