DeMoivre Theorem

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DeMoivre Theorem · Grade 12 · Worksheet 1

1. Use De Moivre's Theorem to find the fourth roots of 81(cos(2π/3) + i sin(2π/3)). Express your answers in rectangular form a + bi. Answer: 3√3/2 + (3/2)i, -3/2 + (3√3/2)i, -3√3/2 - (3/2)i, 3/2 - (3√3/2)i Solution: Identify the parameters. We are finding the fourth roots (n = 4) of 81(cos(2π/3) + i sin(2π/3)). So r = 81 and θ = 2π/3.
   Full step-by-step solution

   Step 1: Identify the parameters. We are finding the fourth roots (n = 4) of 81(cos(2π/3) + i sin(2π/3)). So r = 81 and θ = 2π/3. Step 2: Find the magnitude of the roots: r^(1/4) = 81^(1/4) = 3. Step 3: Apply De Moivre's Theorem for roots. The four roots are: 3[cos((2π/3 + 2πk)/4) + i sin((2π/3 + 2πk)/4)] for k = 0, 1, 2, 3. Step 4: Calculate the angles: For k = 0: θ₀ = (2π/3)/4 = 2π/12 = π/6 For k = 1: θ₁ = (2π/3 + 2π)/4 = (2π/3 + 6π/3)/4 = (8π/3)/4 = 8π/12 = 2π/3 For k = 2: θ₂ = (2π/3 + 4π)/4 = (2π/3 + 12π/3)/4 = (14π/3)/4 = 14π/12 = 7π/6 For k = 3: θ₃ = (2π/3 + 6π)/4 = (2π/3 + 18π/3)/4 = (20π/3)/4 = 20π/12 = 5π/3 Step 5: Write the roots in polar form: Root 1: 3(cos(π/6) + i sin(π/6)) Root 2: 3(cos(2π/3) + i sin(2π/3)) Root 3: 3(cos(7π/6) + i sin(7π/6)) Root 4: 3(cos(5π/3) + i sin(5π/3)) Step 6: Convert each to rectangular form: Root 1: 3(√3/2 + i(1/2)) = (3√3/2) + (3/2)i Root 2: 3(-1/2 + i(√3/2)) = -3/2 + (3√3/2)i Root 3: 3(-√3/2 + i(-1/2)) = -(3√3/2) - (3/2)i Root 4: 3(1/2 + i(-√3/2)) = 3/2 - (3√3/2)i Step 7: The four fourth roots are: (3√3/2) + (3/2)i, -3/2 + (3√3/2)i, -(3√3/2) - (3/2)i, and 3/2 - (3√3/2)i.

2. Find the real part of the complex number (√3 + i)^6 using De Moivre's Theorem. Express your answer as a simplified integer. Answer: -64 Solution: We have \( z = \sqrt{3} + i \). r = \sqrt{(\sqrt{3})^2 + (1)^2} = \sqrt{3 + 1} = \sqrt{4} = 2 We know \( z = r(\cos\theta + i\sin\theta) \). Here, \( \cos\theta = \frac{\sqrt{3}}{2} \) and \( \sin\theta = \frac{1}{2} \).
   Full step-by-step solution

   Let's solve step-by-step using De Moivre's Theorem. --- **Step 1: Write the complex number in polar form** We have \( z = \sqrt{3} + i \). First, find the modulus \( r \): \[ r = \sqrt{(\sqrt{3})^2 + (1)^2} = \sqrt{3 + 1} = \sqrt{4} = 2 \] --- **Step 2: Find the argument \( \theta \)** We know \( z = r(\cos\theta + i\sin\theta) \). Here, \( \cos\theta = \frac{\sqrt{3}}{2} \) and \( \sin\theta = \frac{1}{2} \). From standard angles, \( \theta = \frac{\pi}{6} \) (or 30°). So: \[ z = 2\left( \cos\frac{\pi}{6} + i\sin\frac{\pi}{6} \right) \] --- **Step 3: Apply De Moivre's Theorem** De Moivre's Theorem says: \[ z^n = r^n \left( \cos(n\theta) + i\sin(n\theta) \right) \] Here \( n = 6 \), so: \[ z^6 = 2^6 \left( \cos\left(6 \cdot \frac{\pi}{6}\right) + i\sin\left(6 \cdot \frac{\pi}{6}\right) \right) \] \[ z^6 = 64 \left( \cos\pi + i\sin\pi \right) \] --- **Step 4: Evaluate cosine and sine** \[ \cos\pi = -1, \quad \sin\pi = 0 \] So: \[ z^6 = 64(-1 + i \cdot 0) = -64 \] --- **Step 5: Identify the real part** The number \( z^6 \) is purely real: \( -64 \). Thus, the real part is \( -64 \). --- **Final answer:** -64

3. Noah is a quantum optics researcher studying the behavior of a photon's polarization state in an optical fiber. The polarization state is modeled by the complex number z = 2(cos(π/6) + i sin(π/6)). After the photon passes through 6 identical polarizing filters in sequence, the resulting polarization state is given by z^6. Using De Moivre's Theorem, determine the exact rectangular form of z^6. Answer: -64 Solution: Identify r = 2 and θ = π/6 from the polar form z = 2(cos(π/6) + i sin(π/6)). Apply De Moivre's Theorem: z^6 = [2(cos(π/6) + i sin(π/6))]^6 = 2^6 (cos(6 × π/6) + i sin(6 × π/6)). Compute 2^6 = 64.
   Full step-by-step solution

   Step 1: Identify r = 2 and θ = π/6 from the polar form z = 2(cos(π/6) + i sin(π/6)). Step 2: Apply De Moivre's Theorem: z^6 = [2(cos(π/6) + i sin(π/6))]^6 = 2^6 (cos(6 × π/6) + i sin(6 × π/6)). Step 3: Compute 2^6 = 64. Step 4: Compute the new angle: 6 × π/6 = π. Step 5: Evaluate cos(π) = -1 and sin(π) = 0. Step 6: Substitute: z^6 = 64(-1 + i × 0) = 64 × (-1) = -64. Step 7: The rectangular form is -64 + 0i, which simplifies to -64. The answer is -64.

4. An electrical engineer is analyzing alternating current in a circuit where the voltage is represented by the complex number 2 + 2i. She needs to calculate the voltage after it has been raised to the 6th power to determine power distribution. Using De Moivre's Theorem, find the simplified form of (2 + 2i)^6. Answer: -512i Solution: We have \( z = 2 + 2i \). r = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}. \tan \theta = \frac{2}{2} = 1.
   Full step-by-step solution

   Let's solve step-by-step using De Moivre's Theorem. --- **Step 1: Write the complex number in polar form** We have \( z = 2 + 2i \). First, find the modulus \( r \): \[ r = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}. \] --- **Step 2: Find the argument \( \theta \)** \[ \tan \theta = \frac{2}{2} = 1. \] Since \( 2 + 2i \) is in the first quadrant, \[ \theta = \frac{\pi}{4}. \] So in polar form: \[ z = 2\sqrt{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right). \] --- **Step 3: Apply De Moivre’s Theorem** De Moivre’s Theorem says: \[ z^n = r^n \left( \cos(n\theta) + i \sin(n\theta) \right). \] Here \( n = 6 \): \[ r^6 = (2\sqrt{2})^6. \] --- **Step 4: Compute \( r^6 \)** \[ (2\sqrt{2})^6 = 2^6 \cdot (\sqrt{2})^6. \] \[ 2^6 = 64, \quad (\sqrt{2})^6 = (2^{1/2})^6 = 2^{3} = 8. \] \[ r^6 = 64 \times 8 = 512. \] --- **Step 5: Compute \( n\theta \)** \[ n\theta = 6 \times \frac{\pi}{4} = \frac{6\pi}{4} = \frac{3\pi}{2}. \] --- **Step 6: Apply to trigonometric form** \[ z^6 = 512 \left( \cos \frac{3\pi}{2} + i \sin \frac{3\pi}{2} \right). \] From unit circle: \[ \cos \frac{3\pi}{2} = 0, \quad \sin \frac{3\pi}{2} = -1. \] So: \[ z^6 = 512 \left( 0 + i \cdot (-1) \right) = 512 \cdot (-i) = -512i. \] --- **Final Answer:** \[ -512i \]

5. (2(cos(π/6) + i sin(π/6)))^6 = ? Answer: -64i Solution: De Moivre's theorem provides an efficient way to compute powers of complex numbers. When a complex number is expressed in polar form r(cosθ + i sinθ), raising it to the nth power gives r^n(cos(nθ) + i sin(nθ)).
   Full step-by-step solution

   De Moivre's theorem provides an efficient way to compute powers of complex numbers. When a complex number is expressed in polar form r(cosθ + i sinθ), raising it to the nth power gives r^n(cos(nθ) + i sin(nθ)). This avoids the need for repeated multiplication and simplifies working with complex exponents. The result can then be converted back to rectangular form if needed.

6. Noah is an aerospace engineer designing a satellite's communication system. The signal strength is modeled by the complex number z = 2(cos(π/7) + i sin(π/7)). To analyze the signal after it passes through 7 repeaters in sequence, he needs to compute z^14. Using De Moivre's Theorem, determine the exact rectangular form of z^14. Answer: 16384 Solution: Identify the polar form: z = 2(cos(π/7) + i sin(π/7)) with r = 2 and θ = π/7. Apply De Moivre's Theorem: z^14 = [2(cos(π/7) + i sin(π/7))]^14 = 2^14 (cos(14 × π/7) + i sin(14 × π/7)). Simplify the angle: 14 × π/7 = 2π.
   Full step-by-step solution

   Step 1: Identify the polar form: z = 2(cos(π/7) + i sin(π/7)) with r = 2 and θ = π/7. Step 2: Apply De Moivre's Theorem: z^14 = [2(cos(π/7) + i sin(π/7))]^14 = 2^14 (cos(14 × π/7) + i sin(14 × π/7)). Step 3: Simplify the angle: 14 × π/7 = 2π. Step 4: Evaluate trigonometric functions: cos(2π) = 1, sin(2π) = 0. Step 5: Substitute: z^14 = 2^14 (1 + i × 0) = 2^14. Step 6: Calculate 2^14 = 16384. The answer is 16384.

7. Use De Moivre's Theorem to compute (1 + i)^8 = ? Answer: 16 Solution: Write the complex number in rectangular form: 1 + i Find the modulus: r = sqrt(1^2 + 1^2) = sqrt(2) Find the argument: θ = arctan(1/1) = π/4 Write in polar form: sqrt(2)(cos(π/4) + i sin(π/4)) Apply De Moivre's Theorem: [sqrt(2)(cos(π/4) + i sin(π/4))]^8 = (sqrt(2))^8 (cos(8π/4) + i sin(8π/4))…
   Full step-by-step solution

   Step 1: Write the complex number in rectangular form: 1 + i Step 2: Find the modulus: r = sqrt(1^2 + 1^2) = sqrt(2) Step 3: Find the argument: θ = arctan(1/1) = π/4 Step 4: Write in polar form: sqrt(2)(cos(π/4) + i sin(π/4)) Step 5: Apply De Moivre's Theorem: [sqrt(2)(cos(π/4) + i sin(π/4))]^8 = (sqrt(2))^8 (cos(8π/4) + i sin(8π/4)) Step 6: Simplify: (2^(1/2))^8 = 2^4 = 16 Step 7: Simplify angles: cos(2π) + i sin(2π) = 1 + 0i Step 8: Multiply: 16 × (1 + 0i) = 16 The answer is 16.

8. Use De Moivre's Theorem to find the fourth roots of 81(cos(5π/3) + i sin(5π/3)). Express your answers in rectangular form a + bi. Answer: 3√3/2 - 3/2 i, -3/2 + 3√3/2 i, -3√3/2 + 3/2 i, 3/2 - 3√3/2 i Solution: Identify parameters. r = 81, θ = 5π/3, n = 4. Modulus of roots: 81^(1/4) = 3.
   Full step-by-step solution

   Step 1: Identify parameters. r = 81, θ = 5π/3, n = 4. Step 2: Modulus of roots: 81^(1/4) = 3. Step 3: Angles for k = 0, 1, 2, 3: k = 0: (5π/3 + 0)/4 = 5π/12 k = 1: (5π/3 + 2π)/4 = (5π/3 + 6π/3)/4 = (11π/3)/4 = 11π/12 k = 2: (5π/3 + 4π)/4 = (5π/3 + 12π/3)/4 = (17π/3)/4 = 17π/12 k = 3: (5π/3 + 6π)/4 = (5π/3 + 18π/3)/4 = (23π/3)/4 = 23π/12 Step 4: Roots in polar form: Root 1: 3(cos(5π/12) + i sin(5π/12)) Root 2: 3(cos(11π/12) + i sin(11π/12)) Root 3: 3(cos(17π/12) + i sin(17π/12)) Root 4: 3(cos(23π/12) + i sin(23π/12)) Step 5: Convert to rectangular form using exact values: cos(5π/12) = (√6 - √2)/4, sin(5π/12) = (√6 + √2)/4 → Root 1: 3(√6 - √2)/4 + 3(√6 + √2)/4 i = (3√6 - 3√2)/4 + (3√6 + 3√2)/4 i cos(11π/12) = -(√6 + √2)/4, sin(11π/12) = (√6 - √2)/4 → Root 2: -3(√6 + √2)/4 + 3(√6 - √2)/4 i cos(17π/12) = -(√6 - √2)/4, sin(17π/12) = -(√6 + √2)/4 → Root 3: -3(√6 - √2)/4 - 3(√6 + √2)/4 i cos(23π/12) = (√6 + √2)/4, sin(23π/12) = -(√6 - √2)/4 → Root 4: 3(√6 + √2)/4 - 3(√6 - √2)/4 i The four fourth roots are: (3√6 - 3√2)/4 + (3√6 + 3√2)/4 i, -3(√6 + √2)/4 + 3(√6 - √2)/4 i, -3(√6 - √2)/4 - 3(√6 + √2)/4 i, and 3(√6 + √2)/4 - 3(√6 - √2)/4 i.