Problem 73 - Precalculus Mathematics for Calculus

Question

Find the indicated power using De Moivre’s Theorem. $$\left(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2} i\right)^{12}$$

Step-by-Step Solution

Answer

The result is \(-1\).

Step 1: Express in Polar Form

The complex number \(z = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i\) needs to be converted to polar form. Recall that the modulus \(r\) is given by \(r = \sqrt{\left(\frac{\sqrt{2}}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2} = \sqrt{\frac{1}{2} + \frac{1}{2}} = 1\). The argument \(\theta\) is given by \(\theta = \tan^{-1} \left(\frac{b}{a}\right) = \tan^{-1}(1) = \frac{\pi}{4}\). Thus, \(z\) in polar form is \(1\left(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\right)\).

Step 2: Apply De Moivre's Theorem

According to De Moivre's Theorem, \(z^n = r^n (\cos(n\theta) + i\sin(n\theta))\). Here, \(r = 1\), \(\theta = \frac{\pi}{4}\), and \(n = 12\). Hence, we need to calculate \((1)^{12}(\cos(12 \cdot \frac{\pi}{4}) + i \sin(12 \cdot \frac{\pi}{4}))\).

Step 3: Simplify Argument of Trigonometric Functions

Calculate \(12 \cdot \frac{\pi}{4} = 3\pi\). Knowing that \(\cos(3\pi) = -1\) and \(\sin(3\pi) = 0\), we can substitute these into our equation from Step 2.

Step 4: Calculate Final Result

Substitute the values from Step 3 into the De Moivre's Theorem expression: \(z^{12} = (1)(-1 + 0i) = -1.\)

Key Concepts

Complex Numbers

Complex numbers are fascinating entities that extend the number system beyond real numbers. Each complex number is expressed in the form \(a + bi\), where \(a\) and \(b\) are real numbers and \(i\) is the imaginary unit with the property \(i^2 = -1\). The real part is represented by \(a\), while the imaginary part is \(bi\).

These numbers allow for vast calculations in mathematics and engineering by accommodating non-real solutions.

• The modulus or magnitude of a complex number \(a + bi\) is given by \(r = \sqrt{a^2 + b^2}\).
• The argument or angle \(\theta\) is determined using \(\theta = \tan^{-1}(b/a)\).

Finding the modulus and argument transforms a complex number into polar form, which is a crucial step when dealing with powers and roots of complex numbers using De Moivre's Theorem.

Polar Form

The polar form of a complex number provides a representation that illustrates its geometric properties more naturally. Unlike the rectangular form \(a + bi\), the polar form dissects a complex number into its magnitude and angle relative to the positive real axis.

The polar form is written as \(r (\cos \theta + i \sin \theta)\), where \(r\) is the modulus, and \(\theta\) represents the argument or angle from the real axis.

Why use polar form?

• It simplifies the multiplication and division of complex numbers.
• It makes calculating powers and roots straightforward with De Moivre's Theorem.
• Converting a complex number to polar form can be done using \(r = \sqrt{a^2 + b^2}\) and \(\theta = \tan^{-1}(b/a)\).

Thus, turning \(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i\) into polar form as \(1(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4})\) prepares it for operations like exponentiation as required by De Moivre's Theorem.

Trigonometric Functions

Trigonometric functions like sine and cosine play a vital role in the context of complex numbers. These functions, defined on the unit circle, offer periodic insights into the behavior of angles and help navigate around the circle as transformations are applied.

In the complex plane, the expression \(\cos \theta + i \sin \theta\) naturally follows the circle's path, making it foundational in many mathematical operations. Understanding these functions help in comprehending equations like those involving De Moivre's Theorem,

• By applying these functions to angles as in \(\theta = \frac{\pi}{4}\), a complex number is depicted on the unit circle.
• Periodic properties such as \(\cos 3\pi = -1\) and \(\sin 3\pi = 0\) are crucial in simplifying powers of complex numbers.

These periodic transformations illustrate why calculating \(12 \cdot \frac{\pi}{4} = 3\pi\) and substituting these values leads to the simplified power of \(-1\) as demonstrated by De Moivre's Theorem in the exercise.