Problem 25 - Contemporary Precalculus

Question

Find the nth roots in polar form. $$i ; \quad n=5$$

Step-by-Step Solution

Answer

Answer: The 5th roots of i in polar form are: 1. \(z_0 = \cos(\frac{\pi}{10}) + i\sin(\frac{\pi}{10})\) 2. \(z_1 = \cos(\frac{5\pi}{10}) + i\sin(\frac{5\pi}{10})\) 3. \(z_2 = \cos(\frac{9\pi}{10}) + i\sin(\frac{9\pi}{10})\) 4. \(z_3 = \cos(\frac{13\pi}{10}) + i\sin(\frac{13\pi}{10})\) 5. \(z_4 = \cos(\frac{17\pi}{10}) + i\sin(\frac{17\pi}{10})\)

Step 1: Convert the given complex number to polar form

To convert the complex number i into polar form, we need to find its magnitude and argument(\(\theta\)). The complex number i can be written as x + yi where x = 0 and y = 1. The magnitude (r) is given by: \(r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 1^2} = 1\) The argument(\() \)\theta$ is given by the arctan of the imaginary part over the real part: \(\theta = \arctan(\frac{y}{x}) = \arctan(\frac{1}{0})\) However, the division by zero is undefined. For the imaginary unit i, we know that \(\theta = \frac{\pi}{2}\). So, the polar form of i is: \(1(\cos(\frac{\pi}{2}) + i \sin(\frac{\pi}{2}))\)

Step 2: Apply De Moivre's theorem to find the nth roots of the given complex number in polar form

De Moivre's theorem states that the nth roots of a complex number (z) in polar form (r(cos(\(\theta\)) + isin(\(\theta\)))) are given by the following formula: $$z_k = \sqrt[n]{r}\left(\cos\left(\frac{\theta + 2\pi k}{n}\right) + i \sin\left(\frac{\theta + 2\pi k}{n}\right)\right)$$, where \(k = 0, 1, 2, ..., n-1\). In our case, n = 5 and the complex number is i with r = 1 and \(\theta = \frac{\pi}{2}\). We can plug these values into the formula to find the 5th roots of i: $$z_k = \sqrt[5]{1}\left(\cos\left(\frac{\frac{\pi}{2} + 2\pi k}{5}\right) + i \sin\left(\frac{\frac{\pi}{2} + 2\pi k}{5}\right)\right)$$, where \(k = 0, 1, 2, 3, 4\).

Step 3: Find the 5th roots of i using the formula obtained in step 2

Now, we will find the 5th roots of i using the formula and plugging k = 0, 1, 2, 3, 4: 1. For k = 0: \(z_0 = \cos(\frac{\pi}{10}) + i\sin(\frac{\pi}{10})\) 2. For k = 1: \(z_1 = \cos(\frac{5\pi}{10}) + i\sin(\frac{5\pi}{10})\) 3. For k = 2: \(z_2 = \cos(\frac{9\pi}{10}) + i\sin(\frac{9\pi}{10})\) 4. For k = 3: \(z_3 = \cos(\frac{13\pi}{10}) + i\sin(\frac{13\pi}{10})\) 5. For k = 4: \(z_4 = \cos(\frac{17\pi}{10}) + i\sin(\frac{17\pi}{10})\) These are the 5th roots of the complex number i in polar form.

Key Concepts

Polar Form of Complex Numbers

Understanding the polar form of complex numbers is essential for working with complex roots and powers. Unlike the standard rectangular form, which expresses complex numbers as a sum of a real part and an imaginary part (\( a + bi \)), the polar form focuses on the magnitude and direction of the vector representing the complex number.

Here's a simple breakdown:

• Magnitude (denoted as r) is the distance from the origin to the point in the complex plane.
• Argument (denoted as \( \theta \)) is the angle measured counter-clockwise from the positive x-axis to the vector's terminal point.

The polar form then expresses a complex number as \( r(\cos(\theta) + i\sin(\theta)) \). This form is particularly valuable when multiplying, dividing, or taking powers and roots of complex numbers as it simplifies the calculations.

De Moivre's Theorem

De Moivre's Theorem is a powerful tool for calculating powers and roots of complex numbers. It states that for any complex number \( z = r(\cos(\theta) + i\sin(\theta)) \) and any integer n, the nth power of z is given by:
\[ z^n = r^n(\cos(n\theta)+i\sin(n\theta)) \]
Applying this theorem to find nth roots is a bit trickier. The idea is to take the nth root of the magnitude and divide the argument by n, incorporating multiples of \( 2\pi \) to account for the periodic nature of the trigonometric functions:
\[ z_k = \sqrt[n]{r}\left(\cos\left(\frac{\theta + 2\pi k}{n}\right) + i \sin\left(\frac{\theta + 2\pi k}{n}\right)\right) \]
Here, k represents the kth root (where k is a non-negative integer less than n). By iterating over all possible values of k from 0 to n-1, we obtain all nth roots of the complex number.

Magnitude and Argument of Complex Numbers

Diving into the magnitude and argument of complex numbers enriches our understanding of their properties and is essential for converting to polar form. The magnitude, as previously noted, is the length of the vector starting from the origin to the point representing the complex number. It's calculated using the Pythagorean theorem:
\[ r = \sqrt{x^2 + y^2} \]
where x and y are the real and imaginary parts of the complex number, respectively.

The argument is a bit more nuanced. It can be found by using the arctan (inverse tangent) function when the real part is not zero. When the complex number lies on the imaginary axis (like \( i \)), we refer to known angles and their corresponding points on the unit circle. For instance, \( i \) corresponds to an angle of \( \frac{\pi}{2} \) because it lies at the top of the unit circle.

By knowing the magnitude and argument, we can easily express any complex number in polar form, which opens up a more intuitive way of understanding the complex plane, especially regarding multiplication, division, and finding roots.