Problem 85 - Precalculus with Limits

Question

In Exercises 83-98, (a) use the formula on page 474 to find the indicated roots of the complex number, (b) represent each of the roots graphically, and (c) write each of the root sin standard form. Cube roots of \(8 \left(\cos \dfrac{2\pi}{3}\ +\ i\ \sin \dfrac{2\pi}{3} \right)\)

Step-by-Step Solution

Answer

The cube roots of the complex number are \(2 (\cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3})\), \(2 (\cos 0 + i \sin 0)\) and \(2 (\cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3})\).

Step 1: Recognizing the form

First it's important to recognize that the complex number is already given in polar form. The general form of a complex number in polar coordinates is \(r(\cos \theta + i \sin \theta)\) where \(r\) is the magnitude and \(\theta\) is the argument. So here \(r = 8\) and \(\theta = \frac{2\pi}{3}\).

Step 2: Finding the cube roots

Applying De Moivre's theorem, the cube root of a complex number \(z=r(\cos θ+ i \sin θ)\) will be \(n\) roots as \(z_k=\sqrt[3]{r}(\cos (\frac{θ+2kπ}{3}) + i \sin(\frac{θ+2kπ}{3}))\) where \(k=0, 1, 2\). Therefore, the cube roots are: For \(k=0\), \(z_0 = \sqrt[3]{8} \left( \cos\left(\frac{2\pi}{3}\right) + i \sin\left(\frac{2\pi}{3}\right) \right)\) For \(k=1\), \(z_1 = \sqrt[3]{8} \left( \cos\left(\frac{2\pi + 2\pi}{3}\right) + i \sin\left(\frac{2\pi + 2\pi}{3}\right) \right)\) For \(k=2\), \(z_2 = \sqrt[3]{8} \left( \cos\left(\frac{2\pi + 4\pi}{3}\right) + i \sin\left(\frac{2\pi + 4\pi}{3}\right) \right)\)

Step 3: Simplifying the roots

By calculating the angles and simplifying, the three roots will be: \(z_0 = 2 (\cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3})\) \(z_1 = 2 (\cos 0 + i \sin 0)\) \(z_2 = 2 (\cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3})\)

Key Concepts

Polar Form

Complex numbers can be expressed in different forms including standard and polar. Polar form leverages the magnitude and direction of the vector represented by the complex number. It describes a complex number using its distance from the origin, known as the magnitude (or modulus), and the angle from the positive real axis, known as the argument.
It is represented by \( r(\cos \theta + i \sin \theta) \), where:

• \( r \) is the distance from the origin to the point, akin to the hypotenuse in a right triangle.
• \( \theta \) is the angle made with the real axis, like the angle opposite the hypotenuse.

Understanding this form is a gateway to analyzing complex numbers efficiently, especially when dealing with operations involving multiplication, division, and finding roots. The exercise provided illustrates this by identifying \( r = 8 \) and \( \theta = \frac{2\pi}{3} \), showcasing that polar form simplifies the calculations needed to manipulate complex numbers, especially with angles and roots.

De Moivre's Theorem

De Moivre's theorem is a vital tool in complex number calculations, particularly when raising complex numbers to powers or extracting roots. This theorem states that for any integer \( n \), the power of a complex number in polar form \( (r(\cos \theta + i \sin \theta))^n \) is given by:
\[(r^n)(\cos(n\theta) + i\sin(n\theta))\]This simplifies powering operations because it converts them to simple multiplications in the angle.
In the context of our exercise, De Moivre's theorem helps to find cube roots. Applying this theorem, the cube roots are calculated by adjusting the angle \( \theta \) to \( \frac{(\theta + 2k\pi)}{3} \) while also taking the cube root of \( r \). This results in each root being easily defined by changing \( k \), for values 0, 1, and 2, to yield three distinct roots.

Cube Roots

Finding the cube roots of a complex number involves identifying three separate solutions, all equidistant in the polar plane. The formula we use here derives directly from De Moivre's theorem. The formula is:
\[z_k=\sqrt[3]{r}(\cos (\frac{\theta+2k\pi}{3}) + i \sin(\frac{\theta+2k\pi}{3}))\]for \( k=0,1,2 \).

• For \( k=0 \), the root is positioned at the original argument angle, slightly adjusted for the third root.
• With \( k=1 \), the root is found by increasing the argument angle by \( \frac{2\pi}{3} \) radians or 120 degrees.
• For \( k=2 \), \( 4\pi/3 \) is added to the original angle, placing the root 240 degrees from the original.

Each of these roots has the same magnitude, but different angles in the complex plane. This reveals the beautiful symmetry and cyclical nature of roots in the unit circle.

Standard Form

Standard form of a complex number is expressed simply as \( a + bi \) where \( a \) and \( b \) are real numbers. Converting from polar to standard form involves trigonometric simplification:

• Measure the cosine and sine values of the specified angle \( \theta \).
• Apply these values in the expressions \( z_0, z_1, \) and \( z_2 \) calculated from their polar roots.

For example, in the exercise, when simplified:

• For \( z_0 \), \( \cos \frac{2\pi}{3} = -\frac{1}{2} \) and \( \sin \frac{2\pi}{3} = \frac{\sqrt{3}}{2} \).
• For \( z_1 \), using \( \cos 0 = 1 \) and \( \sin 0 = 0 \) results in the simplest form \( 2 + 0i \).
• For \( z_2 \), applying the cosine and sine of \( \frac{4\pi}{3} \) achieves another distinct solution.

The translations into standard form provide easily interpretable results regarding the real and imaginary parts of these cube roots.