Problem 2 - Complex numbers from A to...Z

Question

Find the cube roots of the following complex numbers: a) \(z=-i\) b) \(z=-27\); c) \(z=2+2 i\) d) \(z=\frac{1}{2}-i \frac{\sqrt{3}}{2}\) e) \(z=18+26 i\)

Step-by-Step Solution

Answer

Question: Find the cube roots of the following complex numbers using De Moivre's theorem and express them in polar form. a) -i b) -27 c) 2 + 2i d) 1/2 - (√3)i/2 e) 18 + 26i Answer: a) Cube roots of -i: z₀ = cos(π/6) - i*sin(π/6) z₁ = cos(5π/6) - i*sin(5π/6) z₂ = cos(3π/2) - i*sin(3π/2) b) Cube roots of -27: z₀ = 3cos(π/3) - 3i*sin(π/3) z₁ = 3cos(π) - 3i*sin(π) z₂ = 3cos(5π/3) - 3i*sin(5π/3) c) Cube roots of 2 + 2i: z₀ = ∛4*cos(π/12) + ∛4*i*sin(π/12) z₁ = ∛4*cos(5π/12) + ∛4*i*sin(5π/12) z₂ = ∛4*cos(9π/12) + ∛4*i*sin(9π/12) d) Cube roots of 1/2 - (√3)i/2: z₀ = cos(π/9) - i*sin(π/9) z₁ = cos(7π/9) - i*sin(7π/9) z₂ = cos(13π/9) - i*sin(13π/9) e) Cube roots of 18 + 26i: z₀ = 2√2*cos(atan(13/9)/3) + 2√2*i*sin(atan(13/9)/3) z₁ = 2√2*cos((atan(13/9)+2π)/3) + 2√2*i*sin((atan(13/9)+2π)/3) z₂ = 2√2*cos((atan(13/9)+4π)/3) + 2√2*i*sin((atan(13/9)+4π)/3)

Step 1: Convert Complex Numbers to Polar Form

To convert a complex number \(z = a + bi\) to polar form, first find its magnitude \(r = \sqrt{a^2 + b^2}\) and its argument \(\theta = \arctan(\frac{b}{a})\). After conversion, the complex number will be in the form \(z = r(\cos(\theta) + i\sin(\theta))\). a) For \(z = -i\), we have \(a = 0\) and \(b = -1\). Therefore, \(r = \sqrt{0^2 + (-1)^2} = 1\) and \(\theta = \arctan(\frac{-1}{0}) = \frac{3\pi}{2}\). So, \(z = \cos(\frac{3\pi}{2}) - i\sin(\frac{3\pi}{2})\). b) For \(z = -27\), we have \(a = -27\) and \(b = 0\). Therefore, \(r = \sqrt{(-27)^2} = 27\) and \(\theta = \arctan(\frac{0}{-27}) = \pi\). So, \(z = 27\cos(\pi) - 27i\sin(\pi)\). c) For \(z = 2 + 2i\), we have \(a = 2\) and \(b = 2\). Therefore, \(r = \sqrt{2^2 + 2^2} = 2\sqrt{2}\) and \(\theta = \arctan(\frac{2}{2}) = \frac{\pi}{4}\). So, \(z = 2\sqrt{2}(\cos(\frac{\pi}{4}) + i\sin(\frac{\pi}{4}))\). d) For \(z = \frac{1}{2} - i\frac{\sqrt{3}}{2}\), we have \(a = \frac{1}{2}\) and \(b = -\frac{\sqrt{3}}{2}\). Therefore, \(r = \sqrt{(\frac{1}{2})^2 + (-\frac{\sqrt{3}}{2})^2} = 1\) and \(\theta = \arctan(\frac{-\sqrt{3}/2}{1/2}) = \frac{5\pi}{3}\). So, \(z = \cos(\frac{5\pi}{3}) - i\sin(\frac{5\pi}{3})\). e) For \(z = 18 + 26i\), we have \(a = 18\) and \(b = 26\). Therefore, \(r = \sqrt{18^2 + 26^2} = 32\) and \(\theta = \arctan(\frac{26}{18}) = \arctan(\frac{13}{9})\). So, \(z = 32\cos(\arctan(\frac{13}{9})) + 32i\sin(\arctan(\frac{13}{9}))\).

Step 2: Find Cube Roots Using De Moivre's Theorem

Now that we have the polar form of the complex numbers, we can find their cube roots using De Moivre's theorem, which states that, if \(z = r(\cos(\theta) + i\sin(\theta))\), then the cube roots of z are given by \(z_{k} = r^{\frac{1}{3}}(\cos(\frac{\theta + 2\pi k}{3}) + i\sin(\frac{\theta + 2\pi k}{3})), k = 0, 1, 2\). Calculate cube roots for each complex number: a) Cube roots of \(z = \cos(\frac{3\pi}{2}) - i\sin(\frac{3\pi}{2})\): Since \(r = 1\), we have \(1^{\frac{1}{3}} = 1\). Then the cube roots are: \(z_0 = \cos(\frac{\frac{3\pi}{2} + 2\pi(0)}{3}) - i\sin(\frac{\frac{3\pi}{2} + 2\pi(0)}{3})\). \(z_1 = \cos(\frac{\frac{3\pi}{2} + 2\pi(1)}{3}) - i\sin(\frac{\frac{3\pi}{2} + 2\pi(1)}{3})\). \(z_2 = \cos(\frac{\frac{3\pi}{2} + 2\pi(2)}{3}) - i\sin(\frac{\frac{3\pi}{2} + 2\pi(2)}{3})\). b) Cube roots of \(z = 27\cos(\pi) - 27i\sin(\pi)\): Since \(r = 27\), we have \(27^{\frac{1}{3}} = 3\). Then the cube roots are: \(z_0 = 3\cos(\frac{\pi + 2\pi(0)}{3}) - 3i\sin(\frac{\pi + 2\pi(0)}{3})\). \(z_1 = 3\cos(\frac{\pi + 2\pi(1)}{3}) - 3i\sin(\frac{\pi + 2\pi(1)}{3})\). \(z_2 = 3\cos(\frac{\pi + 2\pi(2)}{3}) - 3i\sin(\frac{\pi + 2\pi(2)}{3})\). c) Cube roots of \(z = 2\sqrt{2}(\cos(\frac{\pi}{4}) + i\sin(\frac{\pi}{4}))\): Since \(r = 2\sqrt{2}\), we have \((2\sqrt{2})^{\frac{1}{3}} = \sqrt[3]{4}\). Then the cube roots are: \(z_0 = \sqrt[3]{4}\cos(\frac{\frac{\pi}{4} + 2\pi(0)}{3}) + \sqrt[3]{4}i\sin(\frac{\frac{\pi}{4} + 2\pi(0)}{3})\). \(z_1 = \sqrt[3]{4}\cos(\frac{\frac{\pi}{4} + 2\pi(1)}{3}) + \sqrt[3]{4}i\sin(\frac{\frac{\pi}{4} + 2\pi(1)}{3})\). \(z_2 = \sqrt[3]{4}\cos(\frac{\frac{\pi}{4} + 2\pi(2)}{3}) + \sqrt[3]{4}i\sin(\frac{\frac{\pi}{4} + 2\pi(2)}{3})\). d) Cube roots of \(z = \cos(\frac{5\pi}{3}) - i\sin(\frac{5\pi}{3})\): Since \(r = 1\), we have \(1^{\frac{1}{3}} = 1\). Then the cube roots are: \(z_0 = \cos(\frac{\frac{5\pi}{3} + 2\pi(0)}{3}) - i\sin(\frac{\frac{5\pi}{3} + 2\pi(0)}{3})\). \(z_1 = \cos(\frac{\frac{5\pi}{3} + 2\pi(1)}{3}) - i\sin(\frac{\frac{5\pi}{3} + 2\pi(1)}{3})\). \(z_2 = \cos(\frac{\frac{5\pi}{3} + 2\pi(2)}{3}) - i\sin(\frac{\frac{5\pi}{3} + 2\pi(2)}{3})\). e) Cube roots of \(z = 32\cos(\arctan(\frac{13}{9})) + 32i\sin(\arctan(\frac{13}{9}))\): Since \(r = 32\), we have \(32^{\frac{1}{3}} = 2\sqrt{2}\). Then the cube roots are: \(z_0 = 2\sqrt{2}\cos(\frac{\arctan(\frac{13}{9}) + 2\pi(0)}{3}) + 2\sqrt{2}i\sin(\frac{\arctan(\frac{13}{9}) + 2\pi(0)}{3})\). \(z_1 = 2\sqrt{2}\cos(\frac{\arctan(\frac{13}{9}) + 2\pi(1)}{3}) + 2\sqrt{2}i\sin(\frac{\arctan(\frac{13}{9}) + 2\pi(1)}{3})\). \(z_2 = 2\sqrt{2}\cos(\frac{\arctan(\frac{13}{9}) + 2\pi(2)}{3}) + 2\sqrt{2}i\sin(\frac{\arctan(\frac{13}{9}) + 2\pi(2)}{3})\). Now, we have found the cube roots of each complex number.

Key Concepts

Cube Roots

Finding cube roots in the context of complex numbers might seem challenging initially, yet it's an intriguing concept in mathematics education. Essentially, the cube root of a number is a value that, when multiplied by itself twice, returns the original number.

In the realm of complex numbers, this concept holds, but the twist is the involvement of angles and polar coordinates. For a complex number, its cube roots can be explored geometrically on the complex plane. Each has a particular argument or angle, which when multiplied by three yields the original argument of the complex number.

Let's consider the complex number in its polar form, typically expressed as \[ z = r(\cos(\theta) + i\sin(\theta)) \]. To find its cube roots:

• You'll divide the angle by three, yielding three distinct angles.
• Each resulting angle will correspond to one of the cube roots.
• These roots are symmetrically placed on the complex plane, facilitating a rich geometrical understanding.

This symmetry and diversification of angles results in not only varied magnitudes but also rotations, offering a dynamic approach to visualize cube roots.

Polar Form

The Polar form is pivotal in understanding complex numbers, especially when delving into operations like finding roots. Unlike the rectangular form, which uses real and imaginary components, the polar form utilizes a magnitude and an angle. This is expressed as \[ r(\cos(\theta) + i\sin(\theta)) \], where \( r \) is the magnitude, and \( \theta \) is the argument (angle).

This conversion from rectangular to polar form makes it easier to perform multiplication, division, and root operations with complex numbers. Polar form brings in the advantages of:

• Magnitude: Represents the distance from the origin in the complex plane.
• Argument: Denotes the direction or angle from the positive real axis.

By translating complex number operations into polar form, you will find a more intuitive understanding of trigonometric identities and Euler's formula. These connections simplify complex calculations and enhance comprehension in mathematics education.

De Moivre's Theorem

De Moivre's Theorem serves as a bridge connecting complex numbers and trigonometry, making it especially useful in mathematics education. This theorem states that for any complex number in polar form \[ (r(\cos(\theta) + i\sin(\theta))) \] and integer \( n \), raising the complex number to the \( n \)th power yields \[ r^n(\cos(n\theta) + i\sin(n\theta)) \].

It's a cornerstone for calculating powers and roots of complex numbers efficiently. Here's why it's so essential:

• Simplifies computations by transforming multiplicative operations into manageable trigonometric manipulations.
• Facilitates finding \( n \)th roots, such as cube roots, by dividing the angle, \( \theta \), appropriately and adjusting the radial component, \( r \).

Working with this theorem encourages a deeper understanding of rotational symmetry and periodicity in complex numbers. It illustrates the profound interconnectedness between algebra and geometry through transformations on the complex plane.

Mathematics Education

In the context of mathematics education, exploring complex numbers and topics like cube roots, polar forms, and De Moivre's theorem adds an enriching dimension to learning. Students gain:

• Real-World Connections: Understanding these concepts can illuminate applications in physics, engineering, and computer science.
• Enhanced Algebraic Skills: Working with polar forms and roots strengthens algebraic manipulation and problem-solving skills.
• Geometric Insight: Visualizing complex operations on the complex plane contributes to spatial reasoning and geometrical understanding.

A well-rounded approach to teaching these topics helps demystify complex concepts and highlights their beauty in mathematical symmetry and logic. Encouraging explorations in these areas can reveal the underlying simplicity of seemingly intricate mathematical problems.