Question

Find all the complex roots. Write your answers in exponential form. The complex cube roots of -8

Short answer

The complex cube roots of -8 are \(2 \cdot e^{i \frac{ \pi }{3}}, \ 2 \cdot e^{i \pi}, \ 2 \cdot e^{i \frac{ 5 \pi }{3}} \).

Step-by-step solution

Represent in Polar Form

First, express -8 in polar form. -8 can be seen as 8 at an angle of \pi radians (since -8 lies on the negative real axis).So, \(-8 = 8 \cdot e^{i \pi} \).

Apply De Moivre's Theorem

For the complex number in polar form \( r \cdot e^{i \theta} \), De Moivre's theorem states for any integer \ n \, the \ nth \ roots are given by: \( \sqrt[n]{r} \cdot e^{i( \frac{ \theta + 2 \cdot \pi \cdot k}{n})}\), where \ k \) ranges from \( 0 \) to \( n-1 \.

Calculate the Magnitude

For the cube roots of -8, \ n = 3 \, and \ r = 8 \. The magnitude of the cube roots is: \( \sqrt[3]{8} = 2 \).

Determine the Angles

Using \ n = 3 \ and \ \theta = \pi \, calculate the angles for each root: \( \theta_k = \frac{ \pi + 2 \cdot \pi \cdot k}{3} \), \( k = 0, 1, 2 \).

Evaluate Each Root

Plug in values for \( k \):For \ k=0 \: \( \theta_0 = \frac{ \pi }{3} \), providing the root: \( 2 \cdot e^{i \frac{ \pi }{3}} \)For \ k=1 \: \( \theta_1 = \frac{ \pi + 2 \cdot \pi}{3} = \pi \), providing the root: \ 2 \cdot e^{i \pi} \For \ k=2 \: \( \theta_2 = \frac{ \pi + 4 \cdot \pi}{3} = \frac{ 5 \pi }{3} \), providing the root: \( 2 \cdot e^{i \frac{ 5 \pi }{3}} \)

Key concepts

De Moivre's Theorem

De Moivre's Theorem is a powerful tool in complex number theory. It's particularly useful for finding powers and roots of complex numbers. The theorem states that for any complex number in polar form, given by \( r \, \text{e}^{i \theta} \), raised to the power of \( n \), the expression can be written as:
\[ (r \, \text{e}^{i \theta})^n = r^n \, \text{e}^{i n \theta} \]
To find the \( n \)-th roots of a complex number, De Moivre's Theorem helps by transforming the problem into a more manageable form. The \( n \)-th roots are given by:
\[ \sqrt[n]{r} \, \text{e}^{i \frac{ \theta + 2 \pi k }{n}} \]
Where \( k \) ranges from \( 0 \) to \( n-1 \). This results in \( n \) different roots. De Moivre's Theorem also simplifies calculations by making use of angle addition and integer multiples of \( 2\pi \).
To put it simply, De Moivre's theorem connects complex numbers, exponentiation, and trigonometry in a beautiful and functional way, making complex calculations more straightforward.

Polar Form

The polar form of a complex number is one of the ways to represent a complex number, apart from the rectangular (Cartesian) form. A complex number \( z \) can be written as \( z = a + bi \), where \( a \) and \( b \) are real numbers. In polar form, the same number is represented as:
\[ z = r \, \text{e}^{i \theta} \]
Here, \( r \) is the magnitude (or modulus) of the complex number, while \( \theta \) is the argument (or angle). The magnitude \( r \) is the distance from the origin to the point \( (a, b) \) in the complex plane, calculated as:
\[ r = \sqrt{ a^2 + b^2 } \]
The argument \( \theta \) is the angle formed with the positive real axis, calculated using the arctangent function:
\[ \theta = \text{atan2}(b, a) \]
Expressing complex numbers in polar form can simplify multiplication, division, and finding powers and roots of complex numbers. It leverages the angle and magnitude to make these operations more intuitively manageable.

Complex Cube Roots

Finding the cube roots of complex numbers involves determining three unique roots that satisfy the equation \( z^3 = -8 \). Using the earlier polar form and De Moivre's Theorem, we start with:
-8 can be written as \( 8 \, \text{e}^{i \pi} \), the polar form where \( r = 8 \) and \( \theta = \pi \).
To find the cube roots, we use:
\[ \sqrt[3]{8} \, \text{e}^{i \frac{ \pi + 2 \pi k }{3}} \]
First, compute the magnitude:
\( \sqrt[3]{8} = 2 \)
Then, calculate the angles for each root with \( k = 0, 1, 2 \):

• For \( k = 0 \), \( \theta_0 = \frac{ \pi }{3} \). The root is \( 2 \, \text{e}^{i \frac{ \pi }{3}} \).
• For \( k = 1 \), \( \theta_1 = \frac{ \pi + 2 \pi }{3} = \pi \). The root is \( 2 \, \text{e}^{i \pi} \).
• For \( k = 2 \), \( \theta_2 = \frac{ \pi + 4 \pi }{3} = \frac{ 5 \pi }{3} \). The root is \( 2 \, \text{e}^{i \frac{ 5 \pi }{3}} \).

By following these steps, you obtain the three distinct cube roots: \( 2 \, \text{e}^{i \frac{ \pi }{3}} \), \( 2 \, \text{e}^{i \pi} \), and \( 2 \, \text{e}^{i \frac{ 5 \pi }{3}} \). These roots are evenly spaced around the complex plane, reflecting the symmetry in root-finding problems in complex analysis.