Problem 69 - Precalculus Mathematics for Calculus

Question

Find the indicated power using De Moivre’s Theorem. $$(1+i)^{20}$$

Step-by-Step Solution

Answer

The power \((1+i)^{20}\) is \(-1024\).

Step 1: Convert to Polar Form

First, recognize that the number is in the form of a complex number \( z = 1 + i \). We need to find the modulus and argument. The modulus \( r \) is calculated as: \[ r = \sqrt{1^2 + 1^2} = \sqrt{2} \]The argument \( \theta \) is: \[ \theta = \tan^{-1}\left(\frac{1}{1}\right) = \frac{\pi}{4} \]Thus, the polar form is:\[ z = \sqrt{2}\left(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\right) \]

Step 2: Use De Moivre's Theorem

De Moivre’s Theorem states that:\[ (r(\cos \theta + i\sin \theta))^n = r^n (\cos(n\theta) + i\sin(n\theta)) \]For \((1+i)^{20}\), we calculate:\[ r^n = (\sqrt{2})^{20} = 2^{10} = 1024 \]\[ n\theta = 20\times \frac{\pi}{4} = 5\pi \]

Step 3: Simplify Using Periodicity

Recall that cosine and sine have a period of \(2\pi\). Thus,\[ \cos(5\pi) = \cos(\pi) = -1 \]\[ \sin(5\pi) = \sin(\pi) = 0 \]

Step 4: Write the Final Result

Substituting back into De Moivre’s formula gives:\[ (1+i)^{20} = 1024(\cos 5\pi + i\sin 5\pi) = 1024(-1 + 0i) = -1024 \]

Key Concepts

Complex Numbers

Complex numbers combine real numbers with imaginary numbers. The general form of a complex number is written as \( z = a + bi \), where \( a \) is the real part and \( bi \) is the imaginary part. In this formula, \( i \) is the imaginary unit which represents the square root of -1, meaning \( i^2 = -1 \). Complex numbers are incredibly useful in mathematics as they allow for solutions to equations that have no real solution.

• The real part is denoted by \( a \).
• The imaginary part is denoted by \( bi \).
• The absolute value or modulus \( |z| \) is calculated as the square root of the sum of the squares of the real and imaginary parts: \( |z| = \sqrt{a^2 + b^2} \).

The complex number \( z = 1 + i \) combines real part 1 and imaginary part 1, making it a simple yet valuable example. Understanding complex numbers' structure allows us to perform a variety of calculations, including converting them into other forms.

Polar Form

Polar form represents complex numbers in terms of a modulus and an angle, rather than a real and imaginary part. This form makes it easier to multiply and raise complex numbers to powers. The polar form of a complex number \( z = a + bi \) is expressed as \( z = r (\cos \theta + i\sin \theta) \), where \( r \) is the modulus and \( \theta \) is the argument or angle.

• Calculate the modulus \( r \) using: \( r = \sqrt{a^2 + b^2} \).
• Find the angle \( \theta \) by using the arctangent function: \( \theta = \tan^{-1}\left(\frac{b}{a}\right) \).

For the complex number \( 1 + i \), we determine the modulus as \( \sqrt{2} \) and the angle \( \frac{\pi}{4} \). The polar form therefore becomes \( \sqrt{2}\left( \cos\frac{\pi}{4} + i \sin\frac{\pi}{4} \right) \). This transformation to polar form is a crucial step in applying De Moivre's Theorem.

Trigonometric Form of Complex Numbers

The trigonometric form of complex numbers, also known as the polar form, utilizes trigonometric functions to express complex numbers. This form is beneficial when performing operations like raising complex numbers to powers or finding roots. It is given by \( z = r(\cos \theta + i \sin \theta) \).

The utility of the trigonometric form shines in calculations involving exponentiation, such as using De Moivre's Theorem. De Moivre's Theorem states for any integer \( n \),

• \( (r(\cos \theta + i \sin \theta))^n = r^n (\cos n\theta + i \sin n\theta) \)

In solving \( (1+i)^{20} \), we transform \( 1+i \) into its trigonometric form \( \sqrt{2}(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4}) \), use De Moivre's Theorem, and subsequently simplify using the periodic properties of trigonometric functions. This approach delivers the final result efficiently by breaking down the exponentiation process in the complex plane. The method emphasizes the elegance and power of using trigonometric forms for complex numbers.