✨ Level 3 - Topic 3: De Moivre's Theorem - Powers and Roots of Complex Numbers 🔢🔑

1) Introduction to De Moivre's Theorem - Unlocking Powers and Roots 🔑

Welcome to **Level 3 - Topic 3: De Moivre's Theorem**. In this topic, we delve into a powerful theorem in complex number theory – De Moivre's Theorem. This theorem provides a remarkably simple way to calculate powers and roots of complex numbers, especially when they are expressed in polar form. It’s a cornerstone in advanced trigonometry and complex analysis, simplifying calculations that would otherwise be tedious and complex.

What is De Moivre's Theorem?

  • Powers of Complex Numbers: De Moivre's Theorem gives a direct formula for raising a complex number to any integer power.
  • Roots of Complex Numbers: It also extends to finding \( n^{th} \) roots of complex numbers, revealing that every non-zero complex number has exactly \( n \) distinct \( n^{th} \) roots.
  • Connection to Trigonometry and Complex Numbers: The theorem elegantly bridges trigonometry and complex numbers, using trigonometric functions (sine and cosine) to perform algebraic operations on complex numbers.

Why is De Moivre's Theorem Important?

  • Simplifies Complex Calculations: Makes calculating powers and roots of complex numbers straightforward, especially for higher powers and roots.
  • Foundation for Complex Analysis: Essential in many areas of complex analysis, including polynomial factorization, finding solutions to equations, and more.
  • Applications in Engineering and Physics: Used in fields like electrical engineering (AC circuit analysis), signal processing, quantum mechanics, and anywhere complex numbers model periodic phenomena.
  • Elegant Mathematical Result: Demonstrates a beautiful interplay between algebra, trigonometry, and complex numbers.

Topic Overview:

  • Complex Numbers in Polar Form: We'll start by revisiting and reinforcing the polar form representation of complex numbers, which is crucial for De Moivre's Theorem.
  • De Moivre's Theorem for Integer Powers: We'll introduce the theorem and see how to use it to calculate \( z^n \) for a complex number \( z \) and integer \( n \).
  • Finding Roots of Complex Numbers: We'll extend the theorem to find \( n^{th} \) roots of complex numbers, understanding why there are \( n \) distinct roots.
  • Applications and Examples: We'll explore examples of how De Moivre's Theorem is used in various mathematical and applied contexts.

Get ready to unlock the power of De Moivre's Theorem and simplify complex number operations! Let's begin! 🔢🔑

2) Prerequisite: Complex Numbers in Polar Form - A Quick Recap 🔄

De Moivre's Theorem is most effectively applied when complex numbers are in polar form. Let's quickly recap how to represent complex numbers in polar form and convert between rectangular and polar forms.

Definition: Polar Form of a Complex Number

A complex number \( z = x + yi \) can be written in polar form as \( z = r(\cos \theta + i\sin \theta) \) or concisely as \( z = re^{i\theta} \) (Euler's form), where:

  • \( r = |z| = \sqrt{x^2 + y^2} \) is the modulus or magnitude of \( z \). \( r \geq 0 \).
  • \( \theta = \arg(z) \) is the argument of \( z \), the angle that the vector representing \( z \) makes with the positive real axis in the complex plane. \( \theta \) is found using \( \tan \theta = \frac{y}{x} \), considering the quadrant of \( (x, y) \) to get the correct angle. \( \theta \) is not unique; if \( \theta \) is an argument, then \( \theta + 2k\pi \) (for any integer \( k \)) is also an argument. The principal argument is usually chosen in the interval \( (-\pi, \pi] \) or \( [0, 2\pi) \).

Example 1: Converting from Rectangular to Polar Form

Convert the complex number \( z = 1 + i\sqrt{3} \) into polar form \( r(\cos \theta + i\sin \theta) \).

  1. 1. Find the modulus \( r \): \( x = 1 \), \( y = \sqrt{3} \). \( r = |z| = \sqrt{x^2 + y^2} = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \).

    \( r = 2 \)

  2. 2. Find the argument \( \theta \): \( \tan \theta = \frac{y}{x} = \frac{\sqrt{3}}{1} = \sqrt{3} \). Reference angle is \( \arctan(\sqrt{3}) = \frac{\pi}{3} \) (or \( 60^\circ \)). Since \( x = 1 > 0 \) and \( y = \sqrt{3} > 0 \), \( z \) is in the first quadrant. Thus, the principal argument is \( \theta = \frac{\pi}{3} \).

    \( \theta = \frac{\pi}{3} \)

  3. 3. Write in polar form: \( z = r(\cos \theta + i\sin \theta) = 2\left(\cos \frac{\pi}{3} + i\sin \frac{\pi}{3}\right) \).

    \( z = 2\left(\cos \frac{\pi}{3} + i\sin \frac{\pi}{3}\right) \)

Solution: The polar form of \( z = 1 + i\sqrt{3} \) is \( 2\left(\cos \frac{\pi}{3} + i\sin \frac{\pi}{3}\right) \).

Example 2: Converting from Polar to Rectangular Form

Convert the complex number \( z = 3\left(\cos \frac{3\pi}{4} + i\sin \frac{3\pi}{4}\right) \) into rectangular form \( x + yi \).

  1. 1. Identify \( r \) and \( \theta \): From the polar form, \( r = 3 \) and \( \theta = \frac{3\pi}{4} \).
  2. 2. Calculate \( x = r\cos \theta \) and \( y = r\sin \theta \):

    \( x = r\cos \theta = 3\cos \frac{3\pi}{4} = 3 \cdot \left(-\frac{\sqrt{2}}{2}\right) = -\frac{3\sqrt{2}}{2} \).

    \( y = r\sin \theta = 3\sin \frac{3\pi}{4} = 3 \cdot \left(\frac{\sqrt{2}}{2}\right) = \frac{3\sqrt{2}}{2} \).

    \( x = -\frac{3\sqrt{2}}{2}, \quad y = \frac{3\sqrt{2}}{2} \)

  3. 3. Write in rectangular form: \( z = x + yi = -\frac{3\sqrt{2}}{2} + i\frac{3\sqrt{2}}{2} \).

    \( z = -\frac{3\sqrt{2}}{2} + \frac{3\sqrt{2}}{2}i \)

Solution: The rectangular form of \( z = 3\left(\cos \frac{3\pi}{4} + i\sin \frac{3\pi}{4}\right) \) is \( -\frac{3\sqrt{2}}{2} + \frac{3\sqrt{2}}{2}i \).

3) De Moivre's Theorem for Integer Powers - Raising Complex Numbers to Powers 🚀

Now we introduce the main theorem. De Moivre's Theorem provides a formula to calculate integer powers of complex numbers in polar form very efficiently.

De Moivre's Theorem (for Integer Powers)

For any complex number \( z = r(\cos \theta + i\sin \theta) \) and any integer \( n \), the \( n^{th} \) power of \( z \) is given by:

\( z^n = [r(\cos \theta + i\sin \theta)]^n = r^n (\cos(n\theta) + i\sin(n\theta)) \)

In Euler's form: If \( z = re^{i\theta} \), then \( z^n = (re^{i\theta})^n = r^n e^{in\theta} \).

In words: To raise a complex number in polar form to the power \( n \), raise the modulus to the power \( n \) and multiply the argument by \( n \).

Example 3: Calculating Powers using De Moivre's Theorem

Calculate \( (1 + i\sqrt{3})^5 \) using De Moivre's Theorem.

  1. 1. Convert \( z = 1 + i\sqrt{3} \) to polar form: From Example 1, we know \( z = 2\left(\cos \frac{\pi}{3} + i\sin \frac{\pi}{3}\right) \). So, \( r = 2 \) and \( \theta = \frac{\pi}{3} \).
  2. 2. Apply De Moivre's Theorem with \( n = 5 \):

    \( (1 + i\sqrt{3})^5 = \left[2\left(\cos \frac{\pi}{3} + i\sin \frac{\pi}{3}\right)\right]^5 = 2^5 \left(\cos\left(5 \cdot \frac{\pi}{3}\right) + i\sin\left(5 \cdot \frac{\pi}{3}\right)\right) \)

  3. 3. Simplify: \( 2^5 = 32 \). \( 5\cdot \frac{\pi}{3} = \frac{5\pi}{3} \). We need to find \( \cos\left(\frac{5\pi}{3}\right) \) and \( \sin\left(\frac{5\pi}{3}\right) \). \( \frac{5\pi}{3} \) is in the fourth quadrant, coterminal with \( \frac{5\pi}{3} - 2\pi = -\frac{\pi}{3} \) or \( \frac{5\pi}{3} - 2\pi + 2\pi = \frac{5\pi}{3} - 6\pi/3 + 6\pi/3 = \frac{5\pi}{3} \). Reference angle is \( \frac{\pi}{3} \). In QIV, cosine is positive, sine is negative. \( \cos\left(\frac{5\pi}{3}\right) = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \), \( \sin\left(\frac{5\pi}{3}\right) = -\sin\left(\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2} \).

    \( (1 + i\sqrt{3})^5 = 32 \left(\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) = 16 - 16i\sqrt{3} \)

Solution: \( (1 + i\sqrt{3})^5 = 16 - 16i\sqrt{3} \).

Example 4: Power with Negative Integer Exponent

Calculate \( \left[2\left(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\right)\right]^{-3} \) using De Moivre's Theorem.

  1. 1. Apply De Moivre's Theorem with \( n = -3 \): Here \( r = 2 \), \( \theta = \frac{\pi}{4} \), \( n = -3 \).

    \( \left[2\left(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\right)\right]^{-3} = 2^{-3} \left(\cos\left((-3) \cdot \frac{\pi}{4}\right) + i\sin\left((-3) \cdot \frac{\pi}{4}\right)\right) \)

  2. 2. Simplify: \( 2^{-3} = \frac{1}{2^3} = \frac{1}{8} \). \( (-3) \cdot \frac{\pi}{4} = -\frac{3\pi}{4} \). We need \( \cos\left(-\frac{3\pi}{4}\right) \) and \( \sin\left(-\frac{3\pi}{4}\right) \). \( -\frac{3\pi}{4} \) is in the third quadrant. Reference angle \( \frac{3\pi}{4} \) is already in QII, reference is \( \pi - \frac{3\pi}{4} = \frac{\pi}{4} \). For \( -\frac{3\pi}{4} \), reference angle is \( \frac{\pi}{4} \). In QIII, both cosine and sine are negative. \( \cos\left(-\frac{3\pi}{4}\right) = \cos\left(\frac{-3\pi}{4} + 2\pi\right) = \cos\left(\frac{5\pi}{4}\right) = -\cos\left(\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2} \). \( \sin\left(-\frac{3\pi}{4}\right) = -\sin\left(\frac{3\pi}{4}\right) = -\sin\left(\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2} \).

    \( \left[2\left(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\right)\right]^{-3} = \frac{1}{8} \left(-\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}\right) = -\frac{\sqrt{2}}{16} - i\frac{\sqrt{2}}{16} \)

Solution: \( \left[2\left(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\right)\right]^{-3} = -\frac{\sqrt{2}}{16} - i\frac{\sqrt{2}}{16} \).

4) Finding \( n^{th} \) Roots of Complex Numbers - Extending De Moivre's Theorem 🌱

De Moivre's Theorem can be extended to find \( n^{th} \) roots of a complex number. Unlike real numbers, a non-zero complex number has exactly \( n \) distinct \( n^{th} \) roots in the complex plane.

De Moivre's Theorem for \( n^{th} \) Roots

For a complex number \( z = r(\cos \theta + i\sin \theta) \) and a positive integer \( n \), the \( n^{th} \) roots of \( z \) are given by \( n \) distinct complex numbers \( w_k \) for \( k = 0, 1, 2, \ldots, n-1 \), where:

\( w_k = \sqrt[n]{r} \left(\cos\left(\frac{\theta + 2k\pi}{n}\right) + i\sin\left(\frac{\theta + 2k\pi}{n}\right)\right) \)

Or in degrees:

\( w_k = \sqrt[n]{r} \left(\cos\left(\frac{\theta + k \cdot 360^\circ}{n}\right) + i\sin\left(\frac{\theta + k \cdot 360^\circ}{n}\right)\right) \)

For \( k = 0, 1, 2, \ldots, n-1 \). Here, \( \sqrt[n]{r} \) is the principal \( n^{th} \) root of the positive real number \( r \).

Key idea: To find \( n^{th} \) roots, take the \( n^{th} \) root of the modulus and divide the argument by \( n \). To find all distinct roots, add multiples of \( 2\pi \) (or \( 360^\circ \)) to the argument before dividing by \( n \), and use \( k = 0, 1, \ldots, n-1 \) to generate \( n \) distinct roots.

Example 5: Finding Cube Roots of a Complex Number

Find the cube roots of \( z = -8i \).

  1. 1. Convert \( z = -8i \) to polar form: \( z = 0 - 8i \). \( x = 0 \), \( y = -8 \). Modulus \( r = |-8i| = 8 \). Argument \( \theta \): since \( x = 0 \) and \( y = -8 < 0 \), \( z \) is on the negative imaginary axis. So, principal argument \( \theta = -\frac{\pi}{2} \) or \( \frac{3\pi}{2} \) or \( 270^\circ \) or \( -90^\circ \). Let's use \( \theta = \frac{3\pi}{2} \). So, \( z = 8\left(\cos \frac{3\pi}{2} + i\sin \frac{3\pi}{2}\right) \).
  2. 2. Apply De Moivre's Theorem for roots with \( n = 3 \): \( \sqrt[3]{r} = \sqrt[3]{8} = 2 \). The cube roots \( w_k \) for \( k = 0, 1, 2 \) are:

    \( w_k = 2 \left(\cos\left(\frac{\frac{3\pi}{2} + 2k\pi}{3}\right) + i\sin\left(\frac{\frac{3\pi}{2} + 2k\pi}{3}\right)\right) = 2 \left(\cos\left(\frac{3\pi + 4k\pi}{6}\right) + i\sin\left(\frac{3\pi + 4k\pi}{6}\right)\right) \)

  3. 3. Calculate roots for \( k = 0, 1, 2 \):
    • For \( k = 0 \): \( w_0 = 2 \left(\cos\left(\frac{3\pi}{6}\right) + i\sin\left(\frac{3\pi}{6}\right)\right) = 2 \left(\cos\left(\frac{\pi}{2}\right) + i\sin\left(\frac{\pi}{2}\right)\right) = 2(0 + i \cdot 1) = 2i \).
    • For \( k = 1 \): \( w_1 = 2 \left(\cos\left(\frac{3\pi + 4\pi}{6}\right) + i\sin\left(\frac{3\pi + 4\pi}{6}\right)\right) = 2 \left(\cos\left(\frac{7\pi}{6}\right) + i\sin\left(\frac{7\pi}{6}\right)\right) = 2 \left(-\frac{\sqrt{3}}{2} - i\frac{1}{2}\right) = -\sqrt{3} - i \).
    • For \( k = 2 \): \( w_2 = 2 \left(\cos\left(\frac{3\pi + 8\pi}{6}\right) + i\sin\left(\frac{3\pi + 8\pi}{6}\right)\right) = 2 \left(\cos\left(\frac{11\pi}{6}\right) + i\sin\left(\frac{11\pi}{6}\right)\right) = 2 \left(\frac{\sqrt{3}}{2} - i\frac{1}{2}\right) = \sqrt{3} - i \).

    Cube roots are: \( 2i, \quad -\sqrt{3} - i, \quad \sqrt{3} - i \)

Solution: The cube roots of \( -8i \) are \( 2i \), \( -\sqrt{3} - i \), and \( \sqrt{3} - i \).

Example 6: Finding Fourth Roots of Unity

Find the fourth roots of unity (i.e., solve \( w^4 = 1 \)).

  1. 1. Express \( z = 1 \) in polar form: \( z = 1 + 0i \). Modulus \( r = |1| = 1 \). Argument \( \theta = 0 \) (positive real axis). So, \( 1 = 1(\cos 0 + i\sin 0) \).
  2. 2. Apply De Moivre's Theorem for roots with \( n = 4 \): \( \sqrt[4]{r} = \sqrt[4]{1} = 1 \). Fourth roots \( w_k \) for \( k = 0, 1, 2, 3 \) are:

    \( w_k = 1 \cdot \left(\cos\left(\frac{0 + 2k\pi}{4}\right) + i\sin\left(\frac{0 + 2k\pi}{4}\right)\right) = \cos\left(\frac{k\pi}{2}\right) + i\sin\left(\frac{k\pi}{2}\right) \)

  3. 3. Calculate roots for \( k = 0, 1, 2, 3 \):
    • For \( k = 0 \): \( w_0 = \cos(0) + i\sin(0) = 1 + i \cdot 0 = 1 \).
    • For \( k = 1 \): \( w_1 = \cos\left(\frac{\pi}{2}\right) + i\sin\left(\frac{\pi}{2}\right) = 0 + i \cdot 1 = i \).
    • For \( k = 2 \): \( w_2 = \cos\left(\frac{2\pi}{2}\right) + i\sin\left(\frac{2\pi}{2}\right) = \cos(\pi) + i\sin(\pi) = -1 + i \cdot 0 = -1 \).
    • For \( k = 3 \): \( w_3 = \cos\left(\frac{3\pi}{2}\right) + i\sin\left(\frac{3\pi}{2}\right) = 0 + i \cdot (-1) = -i \).

    Fourth roots of unity are: \( 1, \quad i, \quad -1, \quad -i \)

Solution: The fourth roots of unity are \( 1, i, -1, -i \).

5) Application: Deriving Trigonometric Identities 💡

De Moivre's Theorem can be used to derive trigonometric identities for multiple angles. By expanding \( (\cos \theta + i\sin \theta)^n \) using both De Moivre's Theorem and the binomial theorem, and then equating real and imaginary parts, we can obtain identities for \( \cos(n\theta) \) and \( \sin(n\theta) \) in terms of powers of \( \cos \theta \) and \( \sin \theta \).

Example 7: Deriving Double Angle Formulas for Cosine and Sine (n=2)

Use De Moivre's Theorem with \( n = 2 \) to derive identities for \( \cos(2\theta) \) and \( \sin(2\theta) \).

  1. 1. Apply De Moivre's Theorem for \( n = 2 \): \( (\cos \theta + i\sin \theta)^2 = \cos(2\theta) + i\sin(2\theta) \).
  2. 2. Expand \( (\cos \theta + i\sin \theta)^2 \) using binomial theorem:

    \( (\cos \theta + i\sin \theta)^2 = (\cos \theta)^2 + 2(\cos \theta)(i\sin \theta) + (i\sin \theta)^2 \)

    \( = \cos^2 \theta + 2i\cos \theta \sin \theta + i^2 \sin^2 \theta \)

    Since \( i^2 = -1 \), \( = \cos^2 \theta - \sin^2 \theta + i(2\cos \theta \sin \theta) \)

  3. 3. Equate real and imaginary parts:

    Comparing \( \cos(2\theta) + i\sin(2\theta) = (\cos^2 \theta - \sin^2 \theta) + i(2\cos \theta \sin \theta) \), we equate the real parts and the imaginary parts.

    • Real parts: \( \cos(2\theta) = \cos^2 \theta - \sin^2 \theta \) (Double angle formula for cosine)
    • Imaginary parts: \( \sin(2\theta) = 2\cos \theta \sin \theta \) (Double angle formula for sine)

    \( \cos(2\theta) = \cos^2 \theta - \sin^2 \theta \)

    \( \sin(2\theta) = 2\sin \theta \cos \theta \)

Solution: Using De Moivre's Theorem with \( n = 2 \), we derived the double angle formulas for cosine and sine.

6) Practice Questions 🎯

6.1 Fundamental – Powers and Roots using De Moivre's Theorem

1. Convert \( z = \sqrt{3} + i \) into polar form.

2. Convert \( z = -2 - 2i \) into polar form.

3. Convert \( z = 4(\cos(\pi) + i\sin(\pi)) \) into rectangular form.

4. Convert \( z = 2(\cos(\frac{\pi}{6}) + i\sin(\frac{\pi}{6})) \) into rectangular form.

5. Use De Moivre's Theorem to calculate \( (\sqrt{3} + i)^3 \).

6. Use De Moivre's Theorem to calculate \( (-2 - 2i)^4 \).

7. Find the square roots of \( z = 4i \).

8. Find the cube roots of \( z = 8 \).

9. Find the fourth roots of \( z = -16 \).

10. Calculate \( \left[ \cos\left(\frac{\pi}{3}\right) + i\sin\left(\frac{\pi}{3}\right) \right]^6 \) and express in rectangular form.

6.2 Challenging – Advanced Applications and Derivations 💪🚀

1. Use De Moivre's Theorem to derive the triple angle formulas for \( \cos(3\theta) \) and \( \sin(3\theta) \).

2. Find all the fifth roots of \( z = 32(\cos(\frac{5\pi}{6}) + i\sin(\frac{5\pi}{6})) \) and express them in polar form.

3. Solve the equation \( x^3 = 1 + i \) for \( x \) using De Moivre's Theorem to find the roots in polar form.

4. (Conceptual) Explain geometrically why the \( n^{th} \) roots of a complex number are vertices of a regular n-gon in the complex plane.

5. (Proof) Prove De Moivre's Theorem for positive integers \( n \) using mathematical induction. (Bonus challenge).

7) Summary - De Moivre's Theorem - Powers and Roots Unlocked 🎉

  • De Moivre's Theorem: A method to calculate powers and roots of complex numbers in polar form.
  • Polar Form Prerequisite: Complex number must be in polar form \( z = r(\cos \theta + i\sin \theta) = re^{i\theta} \) to apply the theorem effectively.
  • Powers: \( [r(\cos \theta + i\sin \theta)]^n = r^n (\cos(n\theta) + i\sin(n\theta)) \).
  • \( n^{th} \) Roots: \( n \) distinct roots given by \( w_k = \sqrt[n]{r} \left(\cos\left(\frac{\theta + 2k\pi}{n}\right) + i\sin\left(\frac{\theta + 2k\pi}{n}\right)\right) \) for \( k = 0, 1, \ldots, n-1 \).
  • Applications: Simplifying complex number powers, finding roots, deriving trigonometric identities, and more in complex analysis and applied fields.

Congratulations! You've mastered De Moivre's Theorem and learned how to use it to find powers and roots of complex numbers, and even derive trigonometric identities. This theorem is a powerful tool in your advanced trigonometry toolkit and opens doors to further exploration in complex numbers and their applications. Get ready to explore Polar Coordinates and Complex Numbers in the next topic! ✨🔢🔑🚀📐🌟

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