Problem 45 - Precalculus : Building Concepts and Connections

Question

Find all the complex solutions of the equations. $$z^{3}+1=0$$

Step-by-Step Solution

Answer

The complex solutions of the equation \(z^3 + 1 = 0\) are \(z = -1, \frac{1}{2} + i \frac{\sqrt{3}}{2}, \frac{1}{2} - i \frac{\sqrt{3}}{2}\).

Step 1: Rewrite the Equation

Rewrite the equation in the form \(z^3 = -1\). Either the real or the imaginary part of z equals to -1.

Step 2: Calculate the root

Calculate the cube root of -1. The cubic roots of unity are: \(1, -\frac{1}{2} + i \frac{\sqrt{3}}{2}, -\frac{1}{2} - i \frac{\sqrt{3}}{2}\). These roots are often referred to as omega (\(\omega\)) and omega-square (\(\omega^2\)). Plugging these values of omega into -1, we get -1 as -1, \(\frac{1}{2} - i \frac{\sqrt{3}}{2}, \frac{1}{2} + i \frac{\sqrt{3}}{2}\) which are the cube roots of -1.

Step 3: Use De Moivre's Theorem

According to De Moivre's theorem, the three roots of the equation can be obtained by multiplying these values with the cube roots of 1 according to the following formula: \(r(cos\theta + i sin\theta)^n) = r^n(cos(n\theta) + i sin(n\theta))\). In our case, n is 3.

Step 4: Find the roots

Plug these values into the formula and calculate, The solutions are: \(z = -1, \frac{1}{2} + i \frac{\sqrt{3}}{2}, \frac{1}{2} - i \frac{\sqrt{3}}{2}\).

Key Concepts

Cube Roots of Complex Numbers

To understand the cube roots of complex numbers, we first need to grasp what finding a root means. For a cube root, we want to find a number that, when raised to the power of three, equals the original number. In the case of the equation \(z^3 = -1\), we are looking for values of \(z\) that satisfy this equation.

Complex numbers are expressed in the form \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit with \(i^2 = -1\). Finding cube roots in the complex plane involves:

• Expressing the complex number in polar form.
• Calculating the magnitude and angle (argument).
• Using the polar form to find roots.

For the equation \(z^3 = -1\), we can express -1 as a complex number: -1 can be seen as \(1(\cos(\pi) + i\sin(\pi))\), using the polar form. Taking the cube root involves dividing the angle by 3, thereby giving us three solutions evenly spaced around the circle.

Exploring De Moivre's Theorem

De Moivre's Theorem is a powerful tool in finding the roots of complex numbers. It states that for any complex number in polar form \(r(\cos \theta + i \sin \theta)\), raising it to a power \(n\) transforms it to \(r^n(\cos(n\theta) + i \sin(n\theta))\). This theorem helps us calculate powers and roots efficiently.

In our exercise, -1 in polar form is \(1(\cos(\pi) + i \sin(\pi))\). Applying De Moivre’s theorem to find cube roots involves:

• Computing \((-1) = 1(\cos(\pi) + i \sin(\pi))\).
• Finding roots by applying the cube root operation \(((r)^{1/3}(\cos((\theta+2k\pi)/3) + i \sin((\theta+2k\pi)/3)))\).
• Choosing \(k = 0, 1, 2\) for all three cube roots.

Hence, the roots calculated are \(-1, \frac{1}{2} + i \frac{\sqrt{3}}{2}, \frac{1}{2} - i \frac{\sqrt{3}}{2}\).

Roots of Unity and Their Applications

The roots of unity are special solutions to the equation \(z^n = 1\). They are complex numbers evenly spaced on the unit circle in the complex plane. For cube roots of unity, these correspond to the solutions to \(z^3 = 1\):

• \(1\)
• \(-\frac{1}{2} + i \frac{\sqrt{3}}{2}\)
• \(-\frac{1}{2} - i \frac{\sqrt{3}}{2}\)

These roots divide the circle into equal segments of \(120^\circ\). They are highly useful in solving equations and simplifying expressions involving powers of complex numbers. When solving \(z^3 + 1 = 0\), the roots of -1 are derived from multiplying these cube roots of unity by -1.

The concept of roots of unity enables us to explore fundamental properties of numbers and can extend to higher powers, leading to interesting patterns and symmetry properties in mathematics.