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Find the fourth roots of \(-8 i\)

Short Answer

Expert verified
The fourth roots of -8i are approximately: 1, i, -1, and -i.

Step by step solution

01

Convert \(-8i\) to Polar Form

First, let's convert the complex number \(-8i\) into its polar form \(r (\cos θ + i \sin θ)\). We know that \(r = \sqrt{a^2 + b^2}\) and \( θ = \arctan(\frac{b}{a})\), typically, but, since \(a = 0\) and \(b = -8\), we can adjust \(r = |b|\) and \(θ = -\frac{π}{2}\). This gives us \(8 (\cos(-\frac{π}{2}) + i \sin(-\frac{π}{2}))\).
02

Apply De Moivre’s Theorem

We will now apply De Moivre's Theorem to find the four complex roots. We use the formula \( \sqrt[n]{r} ( \cos \frac{θ + 2kπ}{n} + i \sin \frac{θ + 2kπ}{n} )\), where y becomes the fourth root of 8, or 1.5, and n is 4. We perform this operation four times, with the values of k being 0, 1, 2, and 3 respectively.
03

Compute the Roots

The roots will then be: Root 1 for \(k=0: 1.5 (\cos \frac{-\pi / 2 + 2(0)\pi}{4} + i \sin \frac{-\pi / 2 + 2(0)\pi}{4})\), Root 2 for \(k=1: 1.5 (\cos \frac{-\pi / 2 + 2(1)\pi}{4} + i \sin \frac{-\pi / 2 + 2(1)\pi}{4})\), Root 3 for \(k=2: 1.5 (\cos \frac{-\pi / 2 + 2(2)\pi}{4} + i \sin \frac{-\pi / 2 + 2(2)\pi}{4})\), and Root 4 for \(k=3: 1.5 (\cos \frac{-\pi / 2 + 2(3)\pi}{4} + i \sin \frac{-\pi / 2 + 2(3)\pi}{4})\).
04

Simplify the Roots

Simplify the expression of each root by evaluating the trigonometric functions.

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Key Concepts

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

De Moivre's Theorem
Understanding De Moivre's Theorem is key when working with powers and roots of complex numbers. It is a formula that connects complex numbers and trigonometry, enabling us to raise complex numbers to any power with ease.

The theorem states that for a complex number in polar form, that is, in the format of \( r(\cos \theta + i \sin \theta) \), and any integer n, the n-th power of the complex number can be expressed as:
\[ (r(\cos \theta + i \sin \theta))^n = r^n (\cos(n\theta) + i \sin(n\theta)) \]
In the exercise provided, De Moivre's Theorem helps in finding the roots of a complex number. Specifically, when we look for the n-th roots, this theorem can be modified as:
\[ \sqrt[n]{r} ( \cos \frac{\theta + 2k\pi}{n} + i \sin \frac{\theta + 2k\pi}{n} ) \]
Here, the value of \( k \) iterates from 0 to \( n-1 \), giving us the distinct n-th roots of the original complex number.
Polar Form of Complex Numbers
Complex numbers can be expressed in a form that uses a combination of trigonometric functions, known as the polar form. This is incredibly useful when multiplying, dividing, or dealing with powers and roots of complex numbers.

The polar form of a complex number is given by:
\[ r (\cos \theta + i \sin \theta) \]
where \( r \) is the magnitude (or modulus) of the complex number, calculated as \( r = \sqrt{a^2 + b^2} \), and \( \theta \) is the argument (or angle), found by \( \theta = \arctan(\frac{b}{a}) \). However, if the complex number lies along the axes, the angle is determined based on the sign of the imaginary part. In the given exercise, \(-8i\) is purely imaginary and negative, hence \( \theta \) is \( -\frac{\pi}{2} \).

By converting to polar form, complex calculations become simpler as we can directly apply trigonometric identities and properties, as well as De Moivre's Theorem.
Trigonometric Functions
The trigonometric functions, sine (\( \sin \)) and cosine (\( \cos \)), are fundamental when we deal with the polar form of complex numbers and apply De Moivre's Theorem. These functions relate the angles of a right triangle to its sides and have a natural extension to functions on a unit circle.

For any angle \( \theta \), we define:
\[ \cos(\theta) = \frac{\text{adjacent side}}{\text{hypotenuse}} \quad \text{and} \quad \sin(\theta) = \frac{\text{opposite side}}{\text{hypotenuse}} \]
Using these functions, we can analyze and interpret the movement around a circle which is especially useful in expressing the angular component of a complex number in polar form. During the simplification step in the provided exercise, it becomes necessary to evaluate \( \cos \) and \( \sin \) for specific angles to obtain the real and imaginary parts of the roots.

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Most popular questions from this chapter

Round your answers to two decimal places. Lucas pulls a 40 -pound box along a level surface from left to right by attaching a piece of rope to the box and pulling on it with a force \(\mathbf{F}_{1}\) of 20 pounds in the direction \(25^{\circ}\) above the horizontal. A friction force \(\mathbf{F}_{2}\) of 5 pounds is acting on the box as it is being pulled. (A friction force acts in the direction opposite to the direction of motion.) (a) Find the \(x\) and \(y\) components of \(\mathbf{F}_{1}\) (b) Find the \(x\) and \(y\) components of \(\mathbf{F}_{2}\) (c) Use your answers to parts (a) and (b) to express the vector sum \(\mathbf{F}_{1}+\mathbf{F}_{2}\) in terms of its \(x\) and \(y\) components. (d) Give the magnitude and direction of each of the other forces acting on the box. (e) Find the magnitude and direction of the net force acting on the box.

Find the magnitude and direction of each of the given vectors. Express the direction as an angle \(\theta\) in standard position, where \(0^{\circ} \leq \theta<360^{\circ},\) to tuo decimal places. $$\mathbf{v}=-3.2 \mathbf{i}+2 \mathbf{j}$$

Find a unit vector in the same direction as the given vector. $$\mathbf{w}=\langle 1,1\rangle$$

Find \(\mathbf{u}-\mathbf{v}, \mathbf{u}+2 \mathbf{v},\) and \(-3 \mathbf{u}+\mathbf{v}\). $$\mathbf{u}=\left\langle\frac{1}{3}, \frac{2}{5}\right\rangle, \mathbf{v}=\langle 1,2\rangle$$

Round your answers to two decimal places. A golf ball is hit from a tee with a launch angle of \(13.2^{\circ}\) and speed 140 miles per hour. Express the velocity of the ball in component form. (Source: www.golf.com)

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