Problem 63 - A Graphical Approach to College Algebra

Question

Find all \(n\) complex solutions of each equation of the form \(x^{n}=k\) $$x^{3}=-64$$

Step-by-Step Solution

Answer

The complex cube roots of \(-64\) are \(2 + 2i\sqrt{3}\), \(-2 + 2i\sqrt{3}\), and \(-4\).

Step 1: Identify the problem

We want to find all complex numbers that satisfy the equation \(x^3 = -64\). This is a standard problem to find the cube roots of a complex number.

Step 2: Determine the magnitude and argument

Express \(-64\) in polar form. The magnitude \(r\) of \(-64\) is \(64\) since it is 64 units from the origin. The argument \(\theta\) is \(\pi\) (or 180 degrees) because the point \(-64\) lies on the negative real axis.

Step 3: Find the principal root

The principal (or primary) cube root of \(-64\) can be found by taking the cube root of the magnitude and dividing the argument by \(3\). This gives \(r^{1/3} = 4\) and \(\theta/3 = \pi/3\). Therefore, the principal root is \(4\text{cis}(\pi/3)\) which equals \(4 \left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right) = 4 \left(\frac{1}{2} + i \frac{\sqrt{3}}{2}\right) = 2 + 2i\sqrt{3}\).

Step 4: Apply the root formula for all solutions

The general formula for finding all the cube roots is to use \(z_k = r^{1/n} \left( \cos\left( \frac{\theta + 2k\pi}{n} \right) + i\sin\left( \frac{\theta + 2k\pi}{n} \right) \right)\) for \(k=0,1,2\).

Step 5: Calculate all solutions

1. For \(k=0\), \(z_0 = 4 \left(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}\right) = 2 + 2i\sqrt{3}\).2. For \(k=1\), \(z_1 = 4 \left(\cos \left(\frac{\pi}{3} + \frac{2\pi}{3}\right) + i \sin \left(\frac{\pi}{3} + \frac{2\pi}{3}\right)\right) \), which simplifies to \(4(-\frac{1}{2} + i \frac{\sqrt{3}}{2}) = -2 + 2i\sqrt{3}\).3. For \(k=2\), \(z_2 = 4 \left(\cos\left(\frac{\pi}{3} + \frac{4\pi}{3}\right) + i \sin\left(\frac{\pi}{3} + \frac{4\pi}{3}\right)\right) \), which simplifies to \(4(-1) = -4\).

Step 6: Finalize the list of solutions

Thus, the three cube roots of \(-64\) are \(2 + 2i\sqrt{3}\), \(-2 + 2i\sqrt{3}\), and \(-4\).

Key Concepts

Cube Roots

When we talk about cube roots, we're looking at finding all complex numbers that satisfy an equation of the form \(x^3 = k\). If \(k\) is a real number, like in our example where \(k = -64\), we have to convert it into a format called polar form to solve it.To find the cube roots:

• Identify the magnitude, which is the distance from the origin to the point \(k\).
• Find the argument, which tells you the direction of \(k\) from the positive x-axis.

For example, to find the cube roots of \(-64\), first express it in polar form. The three cube roots will be the principal cube root, along with the two other roots calculated using a formula that involves rotating the point in the complex plane to give distinct roots. So for \(-64\), the roots are \(2 + 2i\sqrt{3}\), \(-2 + 2i\sqrt{3}\), and \(-4\). They represent the positions on the complex plane that, when each raised to the third power, result in \(-64\).

Polar Form

Polar form is a way of expressing complex numbers. Instead of using the usual \(a + bi\) form, we write them as \(r(\cos \theta + i \sin \theta)\) or more compactly as \(r\text{cis} \theta\), where:

• \(r\) is the magnitude of the complex number.
• \(\theta\) is the argument or angle.

In our exercise, \(-64\) is converted into polar form to facilitate calculating its cube roots. The magnitude \(r\) is \(64\) because that's how far \(-64\) lies from the origin. The argument \(\theta\) is \(\pi\) because \(-64\) is situated on the negative real axis, equivalent to half a circle around the origin.

Magnitude and Argument

The magnitude and the argument are two essential components in understanding and working with complex numbers in polar form.

Magnitude

The magnitude is like the length of the line from the origin to the point representing the complex number on the complex plane. It's calculated using the formula \(\sqrt{a^2 + b^2}\) for a complex number \(a + bi\). For \(-64\), since it lies entirely along the real axis, the magnitude is \(64\).

Argument

The argument is the angle made with the positive real axis. Measured typically in radians, it helps to orient the complex number on the complex plane. For our example of \(-64\), the argument is \(\pi\), as it exactly lies along the negative real axis, making a half-circle from the positive side.

Complex Solutions

Finding complex solutions involves using the roots of complex numbers. After expressing the original number in polar form, you can find multiple roots by adjusting the argument angle.For each cube root:

• The magnitude is the cube root of the original magnitude.
• The argument is divided by 3, and adjusted by \((2k\pi)/n\) for each root.

In our problem, to find all cube roots of \(-64\), we calculated the primary cube root and then found the other roots by increasing the angle:- For \(k=0\), you get \(2 + 2i\sqrt{3}\).- For \(k=1\), you find \(-2 + 2i\sqrt{3}\) by rotating the root.- For \(k=2\), you get \(-4\), completing a full circle.These are the solutions, which show the power of converting a complex number to polar form: it allows easy computation of any root with the formula \(z_k = r^{1/n} \left( \cos\left( \frac{\theta + 2k\pi}{n} \right) + i\sin\left( \frac{\theta + 2k\pi}{n} \right) \right)\).