De Moivre's Theorem

Hey there, students! 👋 Welcome to one of the most elegant and powerful theorems in mathematics - De Moivre's Theorem! This lesson will transform how you work with complex numbers, especially when dealing with powers and roots. By the end of this lesson, you'll master using De Moivre's theorem to compute powers and nth roots of complex numbers, derive trigonometric identities, and solve complex equations. Get ready to discover why mathematicians consider this theorem a true gem! ✨

Understanding De Moivre's Theorem

Abraham de Moivre, an 18th-century French mathematician, discovered this remarkable theorem that connects complex numbers with trigonometry in the most beautiful way. The theorem states that for any complex number in polar form and any integer n:

If $z = r(\cos θ + i\sin θ)$, then $z^n = r^n(\cos nθ + i\sin nθ)$

This might look simple, but it's incredibly powerful! Let's break this down step by step, students.

First, remember that any complex number can be written in polar form. If you have $z = a + bi$, then:

• $r = |z| = \sqrt{a^2 + b^2}$ (the modulus)
• $θ = \arg(z)$ (the argument, measured from the positive real axis)

So $z = r(\cos θ + i\sin θ)$.

Now here's where De Moivre's magic happens! Instead of multiplying out $(a + bi)^n$ using the binomial theorem (which gets messy very quickly), we can use the polar form and simply multiply the modulus by itself n times and multiply the argument by n.

Let's see this in action with a real example. Consider $z = 1 + i$. To find $z^8$:

• First, convert to polar form: $r = \sqrt{1^2 + 1^2} = \sqrt{2}$ and $θ = \arctan(1/1) = π/4$
• So $z = \sqrt{2}(\cos π/4 + i\sin π/4)$
• Using De Moivre's theorem: $z^8 = (\sqrt{2})^8(\cos 8π/4 + i\sin 8π/4) = 16(\cos 2π + i\sin 2π) = 16(1 + 0i) = 16$

Amazing, right? What would have been a nightmare calculation becomes elegant! 🎯

Finding nth Roots Using De Moivre's Theorem

Now, students, let's explore one of the most practical applications - finding nth roots of complex numbers. This is where De Moivre's theorem really shines!

If $z = r(\cos θ + i\sin θ)$, then the nth roots of z are given by:

$$z^{1/n} = r^{1/n}\left(\cos\frac{θ + 2πk}{n} + i\sin\frac{θ + 2πk}{n}\right)$$

where $k = 0, 1, 2, ..., n-1$.

Notice that we get exactly n different roots! This happens because adding $2π$ to an angle brings us back to the same position, but when we divide by n, we get n distinct positions.

Let's find the cube roots of 8. First, write $8 = 8(\cos 0 + i\sin 0)$.

For $k = 0$: $8^{1/3}(\cos 0/3 + i\sin 0/3) = 2(\cos 0 + i\sin 0) = 2$

For $k = 1$: $8^{1/3}(\cos 2π/3 + i\sin 2π/3) = 2(-1/2 + i\sqrt{3}/2) = -1 + i\sqrt{3}$

For $k = 2$: $8^{1/3}(\cos 4π/3 + i\sin 4π/3) = 2(-1/2 - i\sqrt{3}/2) = -1 - i\sqrt{3}$

These three roots form a beautiful equilateral triangle in the complex plane! This geometric interpretation helps us visualize why there are exactly n nth roots, equally spaced around a circle. 🔄

Deriving Trigonometric Identities

Here's where De Moivre's theorem becomes a powerful tool for discovering trigonometric identities, students! By expanding both sides of the equation and comparing real and imaginary parts, we can derive formulas for $\cos nθ$ and $\sin nθ$.

Let's derive the triple angle formulas. Starting with $(\cos θ + i\sin θ)^3$:

Using De Moivre's theorem: $(\cos θ + i\sin θ)^3 = \cos 3θ + i\sin 3θ$

But we can also expand the left side using the binomial theorem:

$(\cos θ + i\sin θ)^3 = \cos^3 θ + 3i\cos^2 θ\sin θ - 3\cos θ\sin^2 θ - i\sin^3 θ$

Rearranging: $= (\cos^3 θ - 3\cos θ\sin^2 θ) + i(3\cos^2 θ\sin θ - \sin^3 θ)$

Comparing real and imaginary parts:

• $\cos 3θ = \cos^3 θ - 3\cos θ\sin^2 θ = \cos^3 θ - 3\cos θ(1 - \cos^2 θ) = 4\cos^3 θ - 3\cos θ$
• $\sin 3θ = 3\cos^2 θ\sin θ - \sin^3 θ = 3(1 - \sin^2 θ)\sin θ - \sin^3 θ = 3\sin θ - 4\sin^3 θ$

These are the famous triple angle formulas! This method works for any integer n, giving us a systematic way to find multiple angle formulas. 📐

Solving Complex Equations

De Moivre's theorem is also invaluable for solving equations like $z^n = w$, where w is a given complex number. The key insight is that if $z^n = w$, then z is an nth root of w.

Consider solving $z^4 = -16$. First, write $-16$ in polar form:

$-16 = 16(\cos π + i\sin π)$

The fourth roots are:

$$z = 16^{1/4}\left(\cos\frac{π + 2πk}{4} + i\sin\frac{π + 2πk}{4}\right) = 2\left(\cos\frac{π + 2πk}{4} + i\sin\frac{π + 2πk}{4}\right)$$

For $k = 0, 1, 2, 3$:

• $z_1 = 2(\cos π/4 + i\sin π/4) = 2(\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}) = \sqrt{2} + i\sqrt{2}$
• $z_2 = 2(\cos 3π/4 + i\sin 3π/4) = 2(-\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}) = -\sqrt{2} + i\sqrt{2}$
• $z_3 = 2(\cos 5π/4 + i\sin 5π/4) = 2(-\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}) = -\sqrt{2} - i\sqrt{2}$
• $z_4 = 2(\cos 7π/4 + i\sin 7π/4) = 2(\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}) = \sqrt{2} - i\sqrt{2}$

You can verify that each of these solutions satisfies the original equation! 🎯

Conclusion

students, you've just mastered one of the most elegant theorems in mathematics! De Moivre's theorem transforms complex calculations involving powers and roots into simple arithmetic with angles and moduli. Whether you're finding the 10th power of a complex number, extracting cube roots, deriving trigonometric identities, or solving polynomial equations in the complex plane, this theorem is your powerful ally. Remember that the key is converting to polar form first, then applying the theorem systematically. With practice, you'll find that problems that once seemed impossible become routine calculations! 🌟

Study Notes

• De Moivre's Theorem: If $z = r(\cos θ + i\sin θ)$, then $z^n = r^n(\cos nθ + i\sin nθ)$

• nth Roots Formula: $z^{1/n} = r^{1/n}\left(\cos\frac{θ + 2πk}{n} + i\sin\frac{θ + 2πk}{n}\right)$ where $k = 0, 1, 2, ..., n-1$

• Polar Form Conversion: $r = |z| = \sqrt{a^2 + b^2}$ and $θ = \arg(z)$

• Number of nth Roots: Every non-zero complex number has exactly n distinct nth roots

• Geometric Interpretation: nth roots are equally spaced around a circle of radius $r^{1/n}$

• Triple Angle Formulas: $\cos 3θ = 4\cos^3 θ - 3\cos θ$ and $\sin 3θ = 3\sin θ - 4\sin^3 θ$

• Solving $z^n = w$: Find nth roots of w using De Moivre's theorem

• Key Strategy: Always convert to polar form before applying De Moivre's theorem

• Verification: Check answers by substituting back into the original equation