De Moivre's Theorem: nth Power of Complex Numbers

Concept Overview

This question tests the understanding and application of De Moivre's Theorem, a fundamental tool for working with complex numbers in polar form. De Moivre's Theorem provides a direct and elegant method to compute the $n$-th power of a complex number expressed as $r(\cos \theta + i \sin \theta)$. It simplifies complex number exponentiation by relating it to the multiplication of angles.

Worked Solution

Step 1: Understand the Polar Form of a Complex Number A complex number $z$ can be represented in polar form as $z = r(\cos \theta + i \sin \theta)$, where $r$ is the modulus (distance from the origin) and $\theta$ is the argument (angle with the positive real axis). This form is crucial for applying De Moivre's Theorem.

Step 2: State De Moivre's Theorem De Moivre's Theorem states that for any complex number $z = r(\cos \theta + i \sin \theta)$ and any integer $n$, the $n$-th power of $z$ is given by: $z^n = [r(\cos \theta + i \sin \theta)]^n = r^n (\cos(n\theta) + i \sin(n\theta))$ This theorem is the core of solving problems involving powers of complex numbers in polar form.

Step 3: Apply the Theorem to Find the nth Power To find the $n$-th power of a complex number in polar form, follow these steps:

1. Identify the modulus $r$ and the argument $\theta$ of the complex number.
2. Raise the modulus $r$ to the power of $n$, i.e., calculate $r^n$.
3. Multiply the argument $\theta$ by $n$, i.e., calculate $n\theta$.
4. Substitute these values back into the formula from Step 2: $z^n = r^n (\cos(n\theta) + i \sin(n\theta))$.

Step 4: Example Application Let's find the 5th power of the complex number $z = 2(\cos(\frac{\pi}{3}) + i \sin(\frac{\pi}{3}))$. Here, $r = 2$, $\theta = \frac{\pi}{3}$, and $n = 5$.

Using De Moivre's Theorem: $z^5 = 2^5 \left(\cos\left(5 \times \frac{\pi}{3}\right) + i \sin\left(5 \times \frac{\pi}{3}\right)\right)$ $z^5 = 32 \left(\cos\left(\frac{5\pi}{3}\right) + i \sin\left(\frac{5\pi}{3}\right)\right)$ We can further simplify the trigonometric values: $\cos(\frac{5\pi}{3}) = \cos(2\pi - \frac{\pi}{3}) = \cos(\frac{\pi}{3}) = \frac{1}{2}$ $\sin(\frac{5\pi}{3}) = \sin(2\pi - \frac{\pi}{3}) = -\sin(\frac{\pi}{3}) = -\frac{\sqrt{3}}{2}$

So, $z^5 = 32 \left(\frac{1}{2} - i \frac{\sqrt{3}}{2}\right)$ $z^5 = 16 - 16i\sqrt{3}$ This demonstrates how De Moivre's Theorem simplifies the calculation of powers.

Step 5: Connection to Roots of Unity De Moivre's Theorem is also fundamental for finding the $n$-th roots of a complex number. If we want to find the $n$-th roots of unity, we are looking for solutions to $z^n = 1$. In polar form, $1 = 1(\cos(2k\pi) + i \sin(2k\pi))$ for any integer $k$. Applying De Moivre's theorem for roots, the $n$-th roots are given by: $z_k = \sqrt[n]{1} \left(\cos\left(\frac{2k\pi}{n}\right) + i \sin\left(\frac{2k\pi}{n}\right)\right)$ for $k = 0, 1, 2, \dots, n-1$. This shows the theorem's versatility.

Key Takeaways:

• De Moivre's Theorem provides a direct formula for $(r(\cos \theta + i \sin \theta))^n = r^n(\cos(n\theta) + i \sin(n\theta))$.
• It simplifies the calculation of powers of complex numbers by raising the modulus to the power and multiplying the argument by the power.
• The theorem is essential for understanding and calculating roots of unity and other complex number roots.
• Ensure the complex number is in polar form ($r(\cos \theta + i \sin \theta)$) before applying the theorem.

Answer: The $n$-th power of a complex number $z = r(\cos \theta + i \sin \theta)$ is $z^n = r^n(\cos(n\theta) + i \sin(n\theta))$.