Problem 83 - Precalculus Mathematics for Calculus

Question

Find the indicated roots, and graph the roots in the complex plane. The fourth roots of \(-81 i\)

Step-by-Step Solution

Answer

The fourth roots of \(-81i\) are at angles \(-\pi/8, 3\pi/8, 7\pi/8, 11\pi/8\) from the origin.

Step 1: Express in Polar Form

First, we express the complex number \(-81i\) in polar form. A complex number in rectangular form \(a + bi\) can be converted to its polar form \(r(\cos \theta + i\sin \theta)\), where \(r = \sqrt{a^2 + b^2}\) and \(\theta = \tan^{-1}(\frac{b}{a})\).Here, \(a = 0\) and \(b = -81\), which gives:\[r = \sqrt{0^2 + (-81)^2} = 81\]\[\theta = \tan^{-1}(\frac{-81}{0}) = -\frac{\pi}{2}\] (since the number is on the negative imaginary axis).Thus, in polar form, \(\text{-81i} = 81 (\cos(-\frac{\pi}{2}) + i\sin(-\frac{\pi}{2}))\).

Step 2: Apply De Moivre's Theorem

To find the fourth roots, use De Moivre's Theorem, which states that for a complex number in polar form \(r(\cos \theta + i\sin \theta)\), its nth roots are \[r^{1/n} \left(\cos\left(\frac{\theta + 2k\pi}{n}\right) + i\sin\left(\frac{\theta + 2k\pi}{n}\right)\right)\]where \(k = 0, 1, 2, ..., n-1\).For the fourth root, \[n = 4, \quad r^{1/4} = 81^{1/4} = 3\].Thus, the roots are \[3 \left(\cos\left(\frac{-\pi/2 + 2k\pi}{4}\right) + i\sin\left(\frac{-\pi/2 + 2k\pi}{4}\right)\right)\] for \(k = 0, 1, 2, 3\).

Step 3: Calculate Each Root

Now compute each of the four roots.- **For k = 0**: \[3 \left(\cos\left(-\frac{\pi}{8}\right) + i\sin\left(-\frac{\pi}{8}\right)\right)\]- **For k = 1**:\[3 \left(\cos\left(\frac{3\pi}{8}\right) + i\sin\left(\frac{3\pi}{8}\right)\right)\]- **For k = 2**: \[3 \left(\cos\left(\frac{7\pi}{8}\right) + i\sin\left(\frac{7\pi}{8}\right)\right)\]- **For k = 3**: \[3 \left(\cos\left(\frac{11\pi}{8}\right) + i\sin\left(\frac{11\pi}{8}\right)\right)\]

Step 4: Graph the Roots on Complex Plane

Each angle corresponds to a direction on the unit circle in the complex plane. 1. The first root is in the fourth quadrant (slightly below the real axis and right-side). 2. The second root is in the first quadrant. 3. The third root is in the second quadrant. 4. The fourth root is in the third quadrant. Place each of these roots at a distance of 3 from the origin in the direction of their respective angles from Step 3.

Key Concepts

Polar Coordinates

Complex numbers can be represented in two primary forms: rectangular (like the standard form \(a + bi\)) and polar. Understanding polar coordinates is vital as it often simplifies complex number operations, especially when finding roots. In polar form, a complex number is expressed as \(r(\cos \theta + i\sin \theta)\). Here, \(r\) is the modulus, representing the distance of the point from the origin on the complex plane, and \(\theta\) is the argument, indicating the angle with respect to the positive real axis.

To convert a complex number from rectangular to polar form, follow these steps:

• Calculate the modulus \(r\) using \(r = \sqrt{a^2 + b^2}\).
• Calculate the argument \(\theta\) using \(\theta = \tan^{-1}(\frac{b}{a})\).

For the number \(-81i\), the rectangular components are \(a = 0\) and \(b = -81\). Thus, \(r = 81\) and \(\theta = -\frac{\pi}{2}\). This places our number on the negative imaginary axis, or straight down from the real axis. The polar form is then \(81(\cos(-\frac{\pi}{2}) + i\sin(-\frac{\pi}{2}))\).

Polar coordinates make it much easier to apply the powerful mathematical tools like De Moivre's Theorem.

De Moivre's Theorem

De Moivre's Theorem is a cornerstone in complex number calculus, especially useful for raising complex numbers to powers or extracting roots. It elegantly connects powers and roots with polar coordinates.

The theorem states that for a complex number in polar form \(r(\cos \theta + i\sin \theta)\), its nth roots can be found using:

• \(r^{1/n}(\cos(\frac{\theta + 2k\pi}{n}) + i\sin(\frac{\theta + 2k\pi}{n}))\) where \(k\) ranges from \(0\) to \(n-1\).

This formula accounts for all possible nth roots by adjusting the angle \(\theta\) by multiples of \(2\pi\).

For \(-81i\), converting to polar form gives \(r = 81\) and \(\theta = -\frac{\pi}{2}\). Using De Moivre's for the fourth root (\(n = 4\)), we find the roots are:

• \(3(\cos(-\frac{\pi}{8}) + i\sin(-\frac{\pi}{8}))\)
• \(3(\cos(\frac{3\pi}{8}) + i\sin(\frac{3\pi}{8}))\)
• \(3(\cos(\frac{7\pi}{8}) + i\sin(\frac{7\pi}{8}))\)
• \(3(\cos(\frac{11\pi}{8}) + i\sin(\frac{11\pi}{8}))\)

These roots are solutions to the equation \(z^4 = -81i\) in the complex plane.

Complex Plane

The complex plane is an essential visualization tool for understanding complex numbers, akin to how a graph helps with understanding real numbers. It consists of a horizontal real axis and a vertical imaginary axis.

Any complex number \(a + bi\) can be represented as a point on this plane, where \(a\) is the x-coordinate (real part) and \(b\) is the y-coordinate (imaginary part). The modulus \(r\) is the distance from the origin \((0,0)\) to the point, while the argument \(\theta\) is the direction from the positive real axis to the point.

Graphing the fourth roots of \(-81i\) on the complex plane involves placing each root at an angle determined by De Moivre's Theorem:

• The 1st root (when \(k = 0\)) is near the fourth quadrant, slightly below the real axis.
• The 2nd root (when \(k = 1\)) falls in the first quadrant.
• The 3rd root (when \(k = 2\)) appears in the second quadrant.
• The 4th root (when \(k = 3\)) goes into the third quadrant.

Each of these points is \(3\) units from the origin due to their calculated modulus. This spacings help visualize the symmetric and periodic nature of complex number roots on the complex plane.