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Chapter 14: Problem 7

Find the fifth roots of \(-1\), giving the results in polar form. Express the princibal root in the form re \(^{\text {kt }}\),

Short Answer

Expert verified
The fifth roots of \text{-1} are \[ \text{e}^{i\frac{\text{蟺}}{5}}, \text{e}^{i\frac{3\text{蟺}}{5}}, \text{e}^{i\text{蟺}}, \text{e}^{i\frac{7\text{蟺}}{5}}, \text{e}^{i\frac{9\text{蟺}}{5}} \]The principal root is \[ \text{e}^{i\frac{\text{蟺}}{5}} \]

Step by step solution

01

Express \text{-1} in polar form

The number \text{-1} can be written in polar form as \(-1 = \text{e}^{i\text{蟺}}\), where the magnitude is 1 (since \text{-1} lies on the unit circle) and the argument is \text{蟺} (180 degrees).
02

Use De Moivre's Theorem

To find the \text{k}^{th} roots of a complex number, use De Moivre's Theorem: \[ z^{1/n} = r^{1/n} \text{e}^{i(\theta + 2k\text{蟺})/n}\] where k = 0, 1, 2, 3, ..., n-1. Here, \text{n} is 5 for the fifth roots of \text{-1}.
03

Compute the magnitude and argument for the fifth roots

The magnitude r of \text{-1} is 1, and the argument 胃 is 蟺. So, the magnitude of each root will be: \[ r^{1/5} = 1^{1/5} = 1\]The argument for the k^{th} root will be: \[ \frac{\text{蟺} + 2k\text{蟺}}{5} \text{ for } k = 0, 1, 2, 3, 4\]
04

Calculate the arguments for each root

Plugging in values of \text{k} into the formula, we get:For \text{k} = 0: \[ \frac{\text{蟺}}{5}\]For \text{k} = 1: \[ \frac{3\text{蟺}}{5}\]For \text{k} = 2: \[ \frac{5\text{蟺}}{5} = \text{蟺}\]For \text{k} = 3: \[ \frac{7\text{蟺}}{5}\]For \text{k} = 4: \[ \frac{9\text{蟺}}{5}\]
05

Write the fifth roots in polar form

Each fifth root will be: \[ \text{e}^{i \theta_k}\] with \[ \theta_0 = \frac{\text{蟺}}{5} \]\[ \theta_1 = \frac{3\text{蟺}}{5} \]\[ \theta_2 = \text{蟺} \]\[ \theta_3 = \frac{7\text{蟺}}{5} \]\[ \theta_4 = \frac{9\text{蟺}}{5} \]
06

Express the principal root

The principal root is the one with the smallest positive argument, which is: \[ \text{e}^{i\frac{\text{蟺}}{5}}\]

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Key Concepts

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

polar form
Polar form is a way to represent complex numbers. Instead of using Cartesian coordinates (real and imaginary parts), we use magnitude (distance from origin) and argument (angle from the positive x-axis). This is very helpful for problems involving roots and powers of complex numbers. To convert a complex number to polar form:
  • Find the magnitude, which is the distance from the origin. For a number \text{z = x + iy}, the magnitude is given by
    \(r = \sqrt{x^2 + y^2}\).
  • Find the argument, which is the angle the number makes with the positive x-axis. This can be calculated using \(\theta = \tan^{-1}(\frac{y}{x})\) for \text{z = x + iy}.
For example, for \text{-1}, the magnitude is 1 (since it lies on the unit circle) and the argument is \pi (180^\text{degrees}). Therefore, \text{-1} in polar form is \(\text{e}^{i\text{\theta}} = \text{e}^{i\text{\pi}}\).
This setup is useful for applying De Moivre's Theorem to find roots.
De Moivre's Theorem
De Moivre's Theorem is a powerful tool for working with complex numbers in polar form. It links complex numbers and trigonometry through exponential functions. The theorem states:
\[ (\text{r} \text{e}^{i\text{\theta}})^n = r^n \text{e}^{i n \text{\theta}}\] This can be incredibly useful for finding roots of complex numbers. To find the nth root:
  • Take the magnitude \text{r} and raise it to \text{1/n}.
  • Divide the argument \text{\theta} by \text{n}.
  • Adjust the angle by adding intervals of \frac{2\text{\pi}}{n}, covering all possible positive angles.
For the fifth roots of \text{-1}:
  • The magnitude is \1^{1/5} = 1.
  • The primary argument \text{\theta} is \pi.
  • Divide \pi by 5 to get \frac{\pi}{5}.
  • Add \frac{2\text{\pi}k}{5} for \text{k = 0, 1, 2, 3, 4} to get all five roots.
complex numbers
Complex numbers are numbers that have both a real part and an imaginary part. They are written in the form \(z = a + ib\), where \text{a} is the real part and \text{ib} is the imaginary part. Key concepts:
  • Real Part: The component without the imaginary unit, \text{a} in \text{z = a + ib}.
  • Imaginary Part: The component with the imaginary unit, \text{ib}, where \text{i = \sqrt{-1}}.
  • Polar Form: Represented as \text{re}^{i\text{\theta}} or \text{r} \text{cis}\text{\theta}.
Operations using complex numbers:
  • Addition/Subtraction: Combine the real parts and imaginary parts separately.
  • Multiplication: Distribute and use \(i^2 = -1\).
  • Division: Multiply numerator and denominator by the complex conjugate.
Finding roots involves converting to polar form and using De Moivre's Theorem. In the given problem, we converted \text{-1} to polar form and found its fifth roots using these methods.

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Most popular questions from this chapter

Show that the equation \(z^{3}=1\) has one real root and two other roots which are not real, and that, if one of the non-real roots is denoted by w, the other is then \(w^{2}\). Mark on the Argand diagram the points which represent the three roots and show that they are the vertices of an equilateral triangle.

Fxpress \(\cos ^{4} \theta\) in terms of cosines of multiples of \(\theta\).

Find the fifth roots of \(-3+i 3\) in polar form and in exponential form.

Determine the roots of the equation \(x^{3}+64=0\) in the form \(a+j b\), where \(a\) and \(b\) are real.

Find the three cube roots of \(8\left(\cos 264^{\circ}+j \sin 264^{\circ}\right)\) and state which of them is the principal cube root. Show all three roots on an Argand diagram.
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