Problem 59 Find the cube roots of the numbe... [FREE SOLUTION]

Chapter 8: Problem 59

Find the cube roots of the number $$i$$

Short Answer

Expert verified

The cube roots of $i$ are $\text{cos}(\frac{\theta}{8}) + i\text{sin}(\frac{\theta}{8})$, $\text{cos}(2) + i\text{sin}(2)$, and $\text{cos}(\frac{5\theta}{12}) + i\text{sin}(\frac{\theta}{12})$.

Step by step solution

Express the Complex Number in Polar Form

The complex number given is $i$. The polar form of a complex number is given by $z = r(\text{cos}(\theta) + i\text{sin}(\theta))$. For $i$, the modulus $r$ is 1 (since $|i| = 1$), and the argument $\theta$ is $\frac{\theta}{\theta} = \frac{\theta + \theta}{2\theta}$

Use De Moivre's Theorem

De Moivre's Theorem states that $z^n = r^n (\text{cos}(n\theta) + i\text{sin}(n\theta))$. We need to find the cube roots, hence, $z = r^{1/3} (\text{cos}((\theta + 2k\theta)/3) + i\text{sin}((\theta + 2k\theta)/3))$ for $k = 0, 1, 2$. Since $r = 1$, the formula simplifies to $z_k = \text{cos}(\frac{\theta + 2k\theta}{3}) + i\text{sin}(\frac{\theta + 2k\theta}{3})$

Calculate the Argument for Each Root

For $k = 0, \theta_k = \frac{\theta}{3} = \frac{\theta}{8}$; for $k = 1, \theta_1 = \frac{\theta + 2\theta}{3} = 2$; for $k = 2, \theta_2 = \frac{\theta + 4\theta}{3} = \frac{5\theta}{12}$. This means that the cube roots of $i$ are: $z_0 = \text{cos}(\frac{\theta}{8}) + i\text{sin}(\frac{\theta}{8})$, $z_1 = \text{cos}(2) + i\text{sin}(2)$, and $z_2 = \text{cos}(\frac{5\theta}{12}) + i\text{sin}(\frac{\theta}{12})$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

polar form of complex numbers

The polar form of a complex number is an essential concept for understanding various operations on complex numbers. A complex number can be represented in the form of $ z = x + yi $, where $ x $ and $ y $ are real numbers, and $ i $ is the square root of -1. In polar form, the same number is expressed using its modulus (or magnitude) $ r $ and its argument (or angle) $ \theta $. This is written as $ z = r \left( \cos(\theta) + i \sin(\theta) \right) $.

To convert to polar form:

Calculate the modulus using the formula $ r = \sqrt{x^2 + y^2} $.
Determine the argument using $ \theta = \arctan\left( \frac{y}{x} \right) $. Note this can get complex, involving the quadrant of the complex number.

For the complex number $ i $:

The modulus, $ r $, is 1 because $ |i| = 1 $.
The argument, $ \theta $, is $ \frac{\pi}{2} $ because the complex number lies on the positive imaginary axis.

Thus, the polar form of $ i $ is $ 1 \left( \cos \left( \frac{\pi}{2} \right) + i \sin \left( \frac{\pi}{2} \right) \right) $.

De Moivre's Theorem

De Moivre's Theorem is a powerful tool for working with complex numbers, especially when raised to a power or finding roots.
The theorem states: $ \left( r \left( \cos \theta + i \sin \theta \right) \right)^n = r^n \left( \cos \left( n \theta \right) + i \sin \left( n \theta \right) \right) $. This theorem is particularly useful for finding roots of complex numbers.

For cube roots, we use the form:

$ r^{1/3} \left( \cos \left( \frac{\theta + 2k \pi}{3} \right) + i \sin \left( \frac{\theta + 2k \pi}{3} \right) \right) $
where $ k $ takes integer values 0, 1, and 2 to get the different roots.

Applying this to the complex number $ i $:

Modulus, $ r $, remains as 1.
When $ k = 0 $, $ \theta_0 = \frac{\pi/2}{3} = \frac{\pi}{6} $.
When $ k = 1 $, $ \theta_1 = \frac{\pi/2 + 2\pi}{3} = \frac{5\pi}{6} $.
When $ k = 2 $, $ \theta_2 = \frac{\pi/2 + 4\pi}{3} = \frac{9\pi}{6} = \frac{3\pi}{2} $.

This gives us the three roots:

$ \cos \left( \frac{\pi}{6} \right) + i \sin \left( \frac{\pi}{6} \right) $.
$ \cos \left( \frac{5\pi}{6} \right) + i \sin \left( \frac{5\pi}{6} \right) $.
$ \cos \left( \frac{3\pi}{2} \right) + i \sin \left( \frac{3\pi}{2} \right) $.

These are the cube roots of $ i $.

argument of complex numbers

The argument of a complex number is the angle formed with the positive real axis in the complex plane.
It's denoted as $ \theta $ and measured in radians. To find the argument:

Use $ \theta = \arctan \left( \frac{y}{x} \right) $ where $ x $ is the real part and $ y $ is the imaginary part of the complex number $ x + yi $.

Be mindful of the quadrant in which the complex number lies, as $ \theta $ might need adjustment:

If $ z $ is in the first quadrant, $ \theta $ is positive.
If in the second quadrant, $ \theta $ is $ \pi - \arctan \left( \frac{|y|}{|x|} \right) $.
If in the third, $ \theta $ is $ \pi + \arctan \left( \frac{|y|}{|x|} \right) $.
In the fourth, $ \theta $ is $ 2\pi - \arctan \left( \frac{|y|}{|x|} \right) $.

For our example, the complex number $ i $:

The argument is $ \frac{\pi}{2} $ because it lies on the positive imaginary axis.

This angle is utilized in De Moivre's theorem for finding roots and other operations on complex numbers.

91影视

Find the cube roots of the number $$i$$

Short Answer

Step by step solution

Express the Complex Number in Polar Form

Use De Moivre's Theorem

Calculate the Argument for Each Root

Key Concepts

polar form of complex numbers

De Moivre's Theorem

argument of complex numbers

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