Problem 37 - Contemporary Precalculus

Question

Solve the given equation in the complex number system. $$x^{5}-243 i=0$$

Step-by-Step Solution

Answer

Answer: The five complex solutions of the equation are: 1. \(x_1 = 3(\cos(\frac{\pi}{10}) + i\sin(\frac{\pi}{10}))\) 2. \(x_2 = 3(\cos(\frac{5\pi}{10}) + i\sin(\frac{5\pi}{10}))\) 3. \(x_3 = 3(\cos(\frac{9\pi}{10}) + i\sin(\frac{9\pi}{10}))\) 4. \(x_4 = 3(\cos(\frac{13\pi}{10}) + i\sin(\frac{13\pi}{10}))\) 5. \(x_5 = 3(\cos(\frac{17\pi}{10}) + i\sin(\frac{17\pi}{10}))\)

Step 1: Rewrite the equation in polar form

To rewrite the equation in polar form, let \(x = r(\cos(\theta) + i\sin(\theta))\). We can then rewrite the equation as: $$r^5(\cos(5\theta) + i\sin(5\theta)) = 243i$$

Step 2: Apply De Moivre's theorem

Using De Moivre's theorem, we know that: $$z^n = r^n(\cos(n\theta) + i\sin(n\theta))$$ In this case, \(n = 5\), so \(z^5 = r^5(\cos(5\theta) + i\sin(5\theta))\). Comparing this to our equation, we have: \(r^5 = 243\) and \(\sin(5\theta) = 1\), which implies that \(\cos(5\theta) = 0\).

Step 3: Solve for r

To find the magnitude, we can take the fifth root of \(r^5\): $$r = \sqrt[5]{243} = 3$$

Step 4: Solve for θ

To find the angle, we can use \(\sin(5\theta) = 1\) and \(\cos(5\theta) = 0\). Therefore, the possible values of \(5\theta\) are: $$5\theta = (4k+1)\frac{\pi}{2}$$ where \(k = 0, 1, 2, 3, 4\), since we need five distinct values of \(x\). Then, we can find the values of \(\theta\) by dividing by five: $$\theta = \frac{(4k+1)\pi}{10}$$

Step 5: Find the complex solutions

Now, we can substitute the values of \(r\) and \(\theta\) into the polar form of \(x\) and find the five solutions: $$x = 3(\cos(\frac{(4k+1)\pi}{10}) + i\sin(\frac{(4k+1)\pi}{10}))$$ For \(k = 0, 1, 2, 3, 4\), we have the following solutions: $$x_1 = 3(\cos(\frac{\pi}{10}) + i\sin(\frac{\pi}{10}))$$ $$x_2 = 3(\cos(\frac{5\pi}{10}) + i\sin(\frac{5\pi}{10}))$$ $$x_3 = 3(\cos(\frac{9\pi}{10}) + i\sin(\frac{9\pi}{10}))$$ $$x_4 = 3(\cos(\frac{13\pi}{10}) + i\sin(\frac{13\pi}{10}))$$ $$x_5 = 3(\cos(\frac{17\pi}{10}) + i\sin(\frac{17\pi}{10}))$$

Key Concepts

De Moivre's Theorem

De Moivre's Theorem is a powerful tool in dealing with powers and roots of complex numbers. Named after the French mathematician Abraham de Moivre, it provides a straightforward method to raise complex numbers to a power and to find their roots. The theorem states:

• \(z^n = r^n(\cos(n\theta) + i\sin(n\theta))\)

Here, \(z\) is a complex number represented in polar form as \(r(\cos(\theta) + i\sin(\theta))\), where \(r\) is the magnitude of \(z\), and \(\theta\) is the argument, or angle.
This theorem simplifies finding higher powers of complex numbers. In the context of solving equations like the one in the exercise, we use it to express powers and roots in polar form. When the equation involves complex coefficients, like \(243i\) in our problem, converting to polar form allows us to manage the angular components neatly. De Moivre's Theorem elegantly separates the magnitude and direction of complex numbers, allowing computations to proceed smoothly. This is particularly handy when calculating the roots since it neatly slices complex numbers into manageable parts.

Polar Form of Complex Numbers

Complex numbers can be expressed in various forms, Cartesian being the most common, but the polar form provides particular advantages. In polar form, a complex number \(z = a + bi\) is rewritten as \(r(\cos(\theta) + i\sin(\theta))\), often abbreviated as \(r\text{cis}\theta\).

• \(r = \sqrt{a^2 + b^2}\): The "magnitude" or "modulus" of the complex number.
• \(\theta = \text{atan2}(b, a)\): The "argument" or "angle," determined using the \(\text{atan2}\) function.

Expressing complex numbers in polar form is particularly useful when you're dealing with multiplication, division, and, as seen in the exercise, powers and roots of complex numbers.
The conversion to polar form simplifies multiplication and division by turning these operations into additions and subtractions of angles and multiplications/divisions of magnitudes, making complex arithmetic significantly more accessible. This capability is used heavily in De Moivre's Theorem, where the trigonometric form makes handling intricate mathematical expressions more manageable.

Roots of Complex Numbers

Finding the roots of complex numbers can be an intimidating task if approached without the right tools. Polar form and De Moivre's theorem simplify this significantly. To find the \(n\)th root of a complex number, we make use of the property:

• \(z^{1/n} = r^{1/n}\left(\cos\left(\frac{\theta + 2k\pi}{n}\right) + i\sin\left(\frac{\theta + 2k\pi}{n}\right)\right), \text{where } k = 0, 1, 2, ..., n-1\)

This formula indicates that a complex number has \(n\) distinct \(n\)th roots, each at equally spaced angles in the complex plane. For our exercise, where \(n = 5\), the fifth root of \(243i\) is found by first solving for the magnitude \( r \) using the equation \(r^5 = 243\), giving us \(r = 3\).
For the angle part, we derived the angles \( \theta \) by examining the conditions \(\sin(5\theta) = 1\) and \(\cos(5\theta) = 0\), leading to several possible values of \(5\theta\). Finally, dividing by five gives the individual roots' angles, uniquely defining each solution on the complex plane. The systematic combination of these roots reveals the elegant structure and regular spacing inherent to the solution set, demonstrating how complex numbers weave geometry and algebra together.