Problem 82 - Algebra 2

Question

Find each indicated root if it is a real number. $$ \sqrt[4]{-16} $$

Step-by-Step Solution

Answer

The fourth root of -16 is not a real number. The four roots are \(1 + i\), \(-1 + i\), \(-1 - i\), and \(1 - i\).

Step 1: Understand the Problem

You are being asked to find the fourth root of -16. Since the root is even, you should anticipate that the answer will be a complex number.

Step 2: Write the Negative Number in Polar Form

-16 can be written in polar form as \(16e^{i \pi}\). This is due to the fact that in the polar plane, negative numbers on the real axis have an angle of pi.

Step 3: Find the Root

The fourth roots are computed as\(r_k = (16e^{i \pi})^{1/4} = 2e^{i (\frac{\pi}{4} + k*\frac{\pi}{2})}\), where \(k = 0, 1, 2, 3\).

Step 4: Simplify

Substituting \(k = 0, 1, 2, 3\) in the above equation, we get four roots. The roots are\(r_0 = 2e^{i \frac{\pi}{4}} = \sqrt{2} (\cos \frac{\pi}{4} + i \sin \frac{\pi}{4}) = 1 + i\),\(r_1 = 2e^{i \frac{3\pi}{4}} = \sqrt{2} (\cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4}) = -1 + i\),\(r_2 = 2e^{i \frac{5\pi}{4}} = \sqrt{2} (\cos \frac{5\pi}{4} + i \sin \frac{5\pi}{4}) = -1 - i\),\(r_3 = 2e^{i \frac{7\pi}{4}} = \sqrt{2} (\cos \frac{7\pi}{4} + i \sin \frac{7\pi}{4}) = 1 - i\).

Key Concepts

Polar Form

When dealing with complex numbers, especially when taking roots, converting into polar form can simplify calculations greatly. A complex number expressed as a + bi can also be represented in polar form. In polar form, a number is described by its magnitude (distance from the origin), and its angle, known as the argument, from the positive real axis.
To convert a complex number into polar form:

• Calculate the magnitude using the formula \( r = \sqrt{a^2 + b^2} \), where \( a \) and \( b \) are the real and imaginary parts respectively.
• Determine the argument \( \theta \) using \( \theta = \tan^{-1}(\frac{b}{a}) \).

However, for negative real numbers like -16, it's simpler. It's located on the negative x-axis, which means its angle \( \theta = \pi \) radians. Thus, in our problem, \(-16 = 16e^{i\pi}\). Notice how polar form makes the process more intuitive when dealing with non-real solutions.

Roots of Complex Numbers

Finding roots of complex numbers is an intriguing process where polar form becomes essential. To find the \( n \)th root of a complex number expressed in its polar form, if \( z = r e^{i\theta} \), the roots are given by the formula:
\[z_k = r^{1/n} e^{i(\theta + 2k\pi)/n}\]
for \( k = 0, 1, ..., n-1 \). This formula spreads the roots evenly around a circle in the complex plane.
When working through these calculations, remember:

• The number of distinct roots is equal to \( n \), the order of the root you are taking.
• These roots are symmetrically spaced, forming a perfect circular pattern.

Using this method on \(-16\) from our example, which is \( 16e^{i\pi} \) in polar form, the fourth roots involve setting \( n = 4 \) and go through all possible angles by adding \( k\cdot\frac{\pi}{2} \), providing a clear path to discovering all root values.

Even Roots

Even roots of complex numbers present unique characteristics and considerations. For real numbers, whether positive or negative, even roots typically result in complex numbers. This is due to their symmetrical properties.
For example, no real number squared gives \(-16\). Instead, taking even roots relies on the complex plane, where angles matter just as much as magnitudes.

• Evens roots always output a set of complex numbers, except when they are perfectly aligned on the positive real axis.
• The symmetry in even roots results in positive and negative pairs, showing off the "even" pattern in terms of angles.

By understanding complex roots, particularly even roots, you learn that they often provide complete insight into a number's behavior in the complex plane. For \( \sqrt[4]{-16} \), the computed values are neither purely real nor purely imaginary, but express themselves as complex numbers moving in both positive and negative imaginations.