Problem 15 - Contemporary Precalculus

Question

Find the indicated roots of unity and express your answers in the form \(a+b i\). Fourth roots of unity

Step-by-Step Solution

Answer

Answer: The fourth roots of unity are 1, i, -1, and -i.

Step 1: Convert the complex number to polar form

First, we need to express the complex number \(z\) in polar form: \(z=r(\cos{\theta} + i \sin{\theta})\), where \(r\) is the magnitude and \(\theta\) is the argument (angle) of \(z\). In this case, \(z^4=1\). The polar form of \(1\) is \(1(\cos{0} + i\sin{0})\) with magnitude \(r=1\) and argument \(\theta=0\).

Step 2: Apply De Moivre's Theorem

Next, let's apply De Moivre's Theorem for the fourth roots of unity: \(z = \sqrt[4]{1}(\cos{\frac{0+2k\pi}{4}} + i \sin{\frac{0+2k\pi}{4}})\), where \(k=0,1,2,3\) (as there are 4 roots).

Step 3: Find the 4 roots

For each value of \(k\), we will find the corresponding root: 1. For \(k=0\), \(z = \sqrt[4]{1}(\cos{\frac{0}{4}} + i \sin{\frac{0}{4}}) = 1(\cos{0} + i\sin{0}) = \boxed{1}\). 2. For \(k=1\), \(z = \sqrt[4]{1}(\cos{\frac{2\pi}{4}} + i \sin{\frac{2\pi}{4}}) = 1(\cos{\frac{\pi}{2}} + i\sin{\frac{\pi}{2}}) = \boxed{i}\). 3. For \(k=2\), \(z = \sqrt[4]{1}(\cos{\pi} + i \sin{\pi}) = 1(\cos{\pi} + i\sin{\pi}) = \boxed{-1}\). 4. For \(k=3\), \(z = \sqrt[4]{1}(\cos{\frac{6\pi}{4}} + i \sin{\frac{6\pi}{4}}) = 1(\cos{\frac{3\pi}{2}} + i\sin{\frac{3\pi}{2}}) = \boxed{-i}\). So the four roots of unity are \(1\), \(i\), \(-1\), and \(-i\).

Key Concepts

Polar Form of Complex Numbers

Complex numbers can often feel abstract, but expressing them in polar form makes them much easier to understand. Polar form uses the concept of magnitude and angle to represent a complex number.

To understand this, consider a complex number \(z = a + bi\). Here, \(a\) is the real part and \(b\) is the imaginary part. In polar form, this becomes \(z = r(\cos(\theta) + i\sin(\theta))\), where \(r\) is the magnitude of \(z\), and \(\theta\) is the angle the line makes with the positive x-axis.

Here's how you find these:

• The magnitude \(r\) is \( \sqrt{a^2 + b^2} \).
• The angle \(\theta\) (or argument) is found using \( \tan^{-1}(\frac{b}{a}) \).

By expressing a complex number in polar form, we can simplify many complex operations, especially those involving powers and roots.

Understanding De Moivre's Theorem

De Moivre's Theorem is a powerful tool in complex numbers, used primarily for finding powers and roots of complex numbers in polar form. It states that for a complex number \(z = r(\cos\theta + i\sin\theta)\) and an integer \(n\), \(z^n = r^n(\cos(n\theta) + i\sin(n\theta))\).

This theorem is particularly useful when dealing with roots because it simplifies the multiplication and division of angles while retaining the same magnitude transformation.

When applying it to roots, such as fourth roots of unity, we rearrange the equation for \(n\)-th roots: \[z = \sqrt[n]{r}(\cos(\frac{\theta + 2k\pi}{n}) + i\sin(\frac{\theta + 2k\pi}{n}))\]. Here, \(k\) is an integer from 0 to \(n-1\), representing each root uniquely. This allows us to find all distinct roots of a complex number.

Finding Fourth Roots of Unity

Fourth roots of unity are special solutions to the equation \(z^4 = 1\). In mathematical operations involving unity (the number 1), things often simplify beautifully. With roots of unity, we're essentially asking, "What complex numbers, when raised to the fourth power, will result in 1?"

Using polar form, the number 1 is expressed as \(1(\cos 0 + i\sin 0)\) since its magnitude is 1 and its angle is 0. Applying De Moivre's Theorem:

• For \(k=0\): \(z = 1(\cos(\frac{0}{4}) + i\sin(\frac{0}{4})) = 1 \).
• For \(k=1\): \(z = 1(\cos(\frac{\pi}{2}) + i\sin(\frac{\pi}{2})) = i \).
• For \(k=2\): \(z = 1(\cos(\pi) + i\sin(\pi)) = -1 \).
• For \(k=3\): \(z = 1(\cos(\frac{3\pi}{2}) + i\sin(\frac{3\pi}{2})) = -i \).

This results in four distinct fourth roots: \(1\), \(i\), \(-1\), and \(-i\). These roots form a symmetric pattern on the complex plane, illustrating the beauty and balance within mathematical unity.