Problem 32 - Contemporary Precalculus

Question

Solve the given equation in the complex number system. $$x^{6}+64=0$$

Step-by-Step Solution

Answer

Question: Find the six complex roots of the equation \(x^6 = -64\). Answer: The six complex roots of the equation \(x^6 = -64\) are: $$x_0 = 2(\cos(\frac{\pi}{6}) + i\sin(\frac{\pi}{6}))$$ $$x_1 = 2(\cos(\frac{3\pi}{6}) + i\sin(\frac{3\pi}{6}))$$ $$x_2 = 2(\cos(\frac{5\pi}{6}) + i\sin(\frac{5\pi}{6}))$$ $$x_3 = 2(\cos(\frac{7\pi}{6}) + i\sin(\frac{7\pi}{6}))$$ $$x_4 = 2(\cos(\frac{9\pi}{6}) + i\sin(\frac{9\pi}{6}))$$ $$x_5 = 2(\cos(\frac{11\pi}{6}) + i\sin(\frac{11\pi}{6}))$$

Step 1: Rewriting the equation

First, rewrite the equation as follows: $$x^6 = -64$$ Note that \(-64\) can be expressed as \(64cis(\pi)\). We can write the equation as: $$x^6 = 64cis(\pi)$$

Step 2: Taking the 6th root

Now, take the 6th root of both sides of the equation. The left side will give us \(x\), and the right side will give us a complex number in polar form which can be expressed as follows: $$x = \sqrt[6]{64}cis\Big(\frac{\pi + 2k\pi}{6}\Big) = 2cis\Big(\frac{\pi + 2k\pi}{6}\Big)$$ Here, \(k\) can take integer values from 0 to 5 (a total of 6 possible values), which represent the 6 different roots that we need to find.

Step 3: Compute the six roots

Computing the six roots by substituting the values of \(k\) from 0 to 5: For \(k=0\), we have: $$x_0 = 2cis\Big(\frac{\pi}{6}\Big)$$ For \(k=1\), we have: $$x_1 = 2cis\Big(\frac{3\pi}{6}\Big)$$ For \(k=2\), we have: $$x_2 = 2cis\Big(\frac{5\pi}{6}\Big)$$ For \(k=3\), we have: $$x_3 = 2cis\Big(\frac{7\pi}{6}\Big)$$ For \(k=4\), we have: $$x_4 = 2cis\Big(\frac{9\pi}{6}\Big)$$ For \(k=5\), we have: $$x_5 = 2cis\Big(\frac{11\pi}{6}\Big)$$

Step 4: Convert to rectangular form

Using the relationship \(cis(\theta) = \cos\theta + i\sin\theta\), convert each root to rectangular form: $$x_0 = 2(\cos(\frac{\pi}{6}) + i\sin(\frac{\pi}{6}))$$ $$x_1 = 2(\cos(\frac{3\pi}{6}) + i\sin(\frac{3\pi}{6}))$$ $$x_2 = 2(\cos(\frac{5\pi}{6}) + i\sin(\frac{5\pi}{6}))$$ $$x_3 = 2(\cos(\frac{7\pi}{6}) + i\sin(\frac{7\pi}{6}))$$ $$x_4 = 2(\cos(\frac{9\pi}{6}) + i\sin(\frac{9\pi}{6}))$$ $$x_5 = 2(\cos(\frac{11\pi}{6}) + i\sin(\frac{11\pi}{6}))$$

Step 5: Final Answer

These are the six complex roots of the given equation: $$x_0 = 2(\cos(\frac{\pi}{6}) + i\sin(\frac{\pi}{6}))$$ $$x_1 = 2(\cos(\frac{3\pi}{6}) + i\sin(\frac{3\pi}{6}))$$ $$x_2 = 2(\cos(\frac{5\pi}{6}) + i\sin(\frac{5\pi}{6}))$$ $$x_3 = 2(\cos(\frac{7\pi}{6}) + i\sin(\frac{7\pi}{6}))$$ $$x_4 = 2(\cos(\frac{9\pi}{6}) + i\sin(\frac{9\pi}{6}))$$ $$x_5 = 2(\cos(\frac{11\pi}{6}) + i\sin(\frac{11\pi}{6}))$$

Key Concepts

Complex Number System

The complex number system extends the traditional system of real numbers by including imaginary numbers. An imaginary number is defined as the square root of a negative number, with the most fundamental one being the square root of -1, represented as 'i'. Complex numbers are written as a combination of a real part and an imaginary part, in the form of 'a + bi', where 'a' and 'b' are real numbers.

The beauty of complex numbers lies in their ability to accurately represent all possible solutions to polynomial equations, including those with no real solutions, such as the equation provided in our exercise \( x^6 + 64 = 0 \). Unlike real numbers that can only plot a number line, complex numbers have a two-dimensional representation, with the real part 'a' on the x-axis and the imaginary part 'bi' on the y-axis, thus forming a 'complex plane' where each point corresponds to one unique complex number.

Polar Form of Complex Numbers

While complex numbers are commonly expressed in rectangular form as 'a + bi', they can also be represented in polar form. Polar form emphasizes the magnitude and the angle of a complex number. The magnitude is simply the distance from the origin to the point representing the complex number in the complex plane, and the angle (also called the argument) is the counterclockwise angle made with the positive x-axis.

The polar form is written as \( r \cdot \text{cis}(\theta) \) where 'r' stands for magnitude and \( \text{cis}(\theta) \) is shorthand for \( \cos(\theta) + i\sin(\theta) \). This form is particularly useful when solving complex equations like our example, as it simplifies the process of finding roots of complex numbers. The conversion of \( -64 \) to \( 64\text{cis}(\pi) \) in the exercise solution is an application of the polar form in simplifying the calculation of the complex roots.

Roots of Complex Numbers

Solving for the roots of complex numbers often involves raising complex numbers to a power or extracting roots. The nth root of a complex number in polar form can be easily found using the formula: \( r^{1/n} \cdot \text{cis}(\frac{\theta + 2k\pi}{n}) \), where 'n' is the root being taken, and 'k' is an integer that provides the different possible roots (since complex numbers can have multiple nth roots).

In our sample exercise, the sixth roots of -64 are found using this formula, resulting in six distinct solutions. Each value of 'k' from 0 to 5 gives a different root due to the periodic nature of trigonometric functions, effectively dividing the circle into six equal parts and locating each root at those divisions. This demonstrates the fascinating capability of the complex number system to provide a more comprehensive set of solutions compared to real numbers alone.