Question

Follow the procedure illustrated in Example 4 to determine the indicated roots of the given complex number. $$ 1^{1 / 4} $$

Short answer

Answer: The fourth roots of the complex number 1 are: 1. \(1(\cos(0) + i\sin(0))\). 2. \(1(\cos(\frac{1}{2}) + i\sin(\frac{1}{2}))\). 3. \(1(\cos(1) + i\sin(1))\). 4. \(1(\cos(\frac{3}{2}) + i\sin(\frac{3}{2}))\).

Step-by-step solution

Convert the given complex number to polar form

To find the roots, we first need to convert the given complex number to its polar form. The complex number is 1, which can be represented as \(1 + 0i\). In polar form, we represent a complex number as \(r(\cos(\theta) + i\sin(\theta))\), where \(r\) is the magnitude and \(\theta\) is the angle. The magnitude (r) can be calculated using the formula \(r =\sqrt{a^2 + b^2}\), where \(a\) and \(b\) are the real and imaginary parts of the complex number, respectively. In this case, \(a = 1\) and \(b = 0\). $$ r = \sqrt{1^2 + 0^2} = \sqrt{1} = 1 $$ The angle \((\theta)\) can be calculated using the formula \(\theta = \arctan(\frac{b}{a})\). In this case, since \(a = 1\) and \(b = 0\), the angle is: $$ \theta = \arctan(\frac{0}{1}) = 0 $$ So, the polar form of the given complex number is \(1(\cos(0) + i\sin(0))\).

Use De Moivre's theorem to find the fourth roots

According to De Moivre's theorem, to find the n-th root of a complex number in polar form, we should divide the angle \((\theta)\) by n and raise the magnitude \((r)\) to the power of \(\frac{1}{n}\). In this case, we want to find the fourth roots, so \(n = 4\). For each root, the angle will be: $$ \theta_k = \frac{(2k + 0)}{4} = \frac{2k}{4}, \qquad k = 0, 1, 2, 3 $$ Also, the magnitude of the roots will be: $$ r_k = (1)^{\frac{1}{4}} = 1 $$ Now we can find the fourth roots: 1. \(k = 0\): \(\theta_0 = \frac{2\cdot 0}{4} = 0\), so the root is \(1(\cos(0) + i\sin(0))\). 2. \(k = 1\): \(\theta_1 = \frac{2\cdot 1}{4} = \frac{1}{2}\), so the root is \(1(\cos(\frac{1}{2}) + i\sin(\frac{1}{2}))\). 3. \(k = 2\): \(\theta_2 = \frac{2\cdot 2}{4} = 1\), so the root is \(1(\cos(1) + i\sin(1))\). 4. \(k = 3\): \(\theta_3 = \frac{2\cdot 3}{4} = \frac{3}{2}\), so the root is \(1(\cos(\frac{3}{2}) + i\sin(\frac{3}{2}))\).

Writing the roots in rectangular form (optional)

If you need the roots in rectangular form (a + bi), you can use the trigonometric identities: $$ \cos(\frac{n\pi}{4}) = \frac{a}{\sqrt{1 + a^2}} $$ and $$ \sin(\frac{n\pi}{4}) = \frac{a\sqrt{1 - a^2}}{1+a^2}. $$ However, since the question doesn't specifically ask for the roots in rectangular form, we'll leave them in polar form. So, the fourth roots of the given complex number 1 are: 1. \(1(\cos(0) + i\sin(0))\). 2. \(1(\cos(\frac{1}{2}) + i\sin(\frac{1}{2}))\). 3. \(1(\cos(1) + i\sin(1))\). 4. \(1(\cos(\frac{3}{2}) + i\sin(\frac{3}{2}))\).

Key concepts

Polar Form

To understand complex numbers fully, it's important to know how to convert them into polar form. A complex number like 1 can be expressed as 1 + 0i in its basic rectangular form. In polar terms, it's represented as \( r(\cos(\theta) + i\sin(\theta)) \). Here, \( r \) is the magnitude and \( \theta \) is the angle (often called the argument).
The magnitude \( r \) is found using the formula \( r = \sqrt{a^2 + b^2} \), where \( a \) is the real part, and \( b \) is the imaginary part of the number.
For the number 1, since \( a = 1 \) and \( b = 0 \), we get \( r = \sqrt{1^2 + 0^2} = 1 \).
The angle \( \theta \) is calculated as \( \theta = \arctan(\frac{b}{a}) \). Here, that means \( \theta = \arctan(\frac{0}{1}) = 0 \).

• The polar form of 1 is thus \( 1(\cos(0) + i\sin(0)) \).

This shows how a number in geometric form gives intuitive insight into its size and direction.

De Moivre's Theorem

De Moivre's theorem is a powerful tool that simplifies many operations on complex numbers when they are in polar form. It's used to find powers and roots of complex numbers.
According to this theorem, if you have a complex number in the form of \( r(\cos(\theta) + i\sin(\theta)) \), its \( n \)-th root is determined by:

• New magnitude: \( r^{1/n} \)
• New angles: \( \frac{\theta + 2k\pi}{n} \), where \( k \) is an integer.

The theorem works because of the periodic nature of trigonometric functions, making it possible to rotate the angle sufficiently to get all necessary roots.
In our example, we find the fourth roots of 1 by dividing the angle by 4 while maintaining a magnitude of 1. We compute four roots by substituting \( k = 0, 1, 2, 3 \).

Fourth Roots

To understand the fourth roots of a complex number, such as 1, one can use both geometric and algebraic perspectives. The fourth roots, viewed geometrically, are points evenly spaced on a circle in the complex plane.
For a complex number of magnitude 1 like in this example, every root has a magnitude of \( 1^{1/4} = 1 \).
What varies is the angle: we compute four distinct angles by using \( \theta_k = \frac{2k\pi}{4} \) for \( k = 0, 1, 2, 3 \).

• \( k = 0 \) gives angle \( 0 \). Root is \( 1(\cos(0) + i\sin(0)) \).
• \( k = 1 \) gives angle \( \frac{\pi}{2} \). Root is \( 1(\cos(\frac{\pi}{2}) + i\sin(\frac{\pi}{2})) \).
• \( k = 2 \) gives angle \( \pi \). Root is \( 1(\cos(\pi) + i\sin(\pi)) \).
• \( k = 3 \) gives angle \( \frac{3\pi}{2} \). Root is \( 1(\cos(\frac{3\pi}{2}) + i\sin(\frac{3\pi}{2})) \).

These angles demonstrate the calculation of roots in their pure polar form.

Rectangular Form

Complex numbers are often represented in what's called rectangular form, which is \( a + bi \). To shift from polar to rectangular form, we apply trigonometric functions to get the Cartesian coordinates \( a \) and \( b \).
For a complex number \( r(\cos(\theta) + i\sin(\theta)) \), we find it in rectangular form by calculating:

• \( a = r\cos(\theta) \)
• \( b = r\sin(\theta) \)

This converts each point in polar coordinates into a point on the complex plane.
For instance, the polar root \( 1(\cos(\frac{\pi}{2}) + i\sin(\frac{\pi}{2})) \) translates to \( 0 + i \cdot 1 \), or just \( i \) in rectangular form.
This process helps in visualizing and performing arithmetic on complex roots.