/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 28 In Exercises 23 through 28 find ... [FREE SOLUTION] | 91影视

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Chapter 0: Problem 28

In Exercises 23 through 28 find all the solutions of the given equations. \(z^{3}=-125 i\)

Short Answer

Expert verified
The solutions are \(5\left(\cos(-\frac{\pi}{6}) + i\sin(-\frac{\pi}{6})\right)\), \(5\left(\cos(\frac{\pi}{2}) + i\sin(\frac{\pi}{2})\right)\), and \(5\left(\cos(\frac{7\pi}{6}) + i\sin(\frac{7\pi}{6})\right)\).

Step by step solution

01

Express in Polar Form

First, express the complex number \(-125i\)in polar form. Recognize that \(-125i\)is equivalent to \(125e^{-\frac{\pi}{2}i}\) because its magnitude \(r=125\) and argument \(-\frac{\pi}{2}\) (sine gives -1).
02

Identify the Roots

Since we need to find the cube roots of \(125e^{-\frac{\pi}{2}i}\), set \(w = 125^{1/3} e^{-\frac{\pi}{6}i}\)\( = 5 e^{-\frac{\pi}{6}i}\).
03

Apply De Moivre's Theorem

Using De Moivre's Theorem, find all the cube roots:\[z_k = 5 \cdot e^{i\left(-\frac{\pi}{6} + \frac{2\pi k}{3}\right)} \text{ for } k = 0, 1, 2.\]
04

Calculate Each Root

For each integer \(k = 0, 1, 2\), calculate:1. For \(k = 0\): \[z_0 = 5e^{-\frac{\pi}{6}i} = 5 \left( \cos\left(-\frac{\pi}{6}\right) + i \sin\left(-\frac{\pi}{6}\right) \right)\].2. For \(k = 1\): \[z_1 = 5e^{i\frac{\pi}{2}} = 5 \left( \cos\left(\frac{\pi}{2}\right) + i \sin\left(\frac{\pi}{2}\right) \right)\].3. For \(k = 2\): \[z_2 = 5e^{i\frac{7\pi}{6}} = 5 \left( \cos\left(\frac{7\pi}{6}\right) + i \sin\left(\frac{7\pi}{6}\right) \right)\].

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Key Concepts

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

Polar Form
The polar form of a complex number is a different way to express the number, which can be particularly useful in multiplying and finding powers and roots of complex numbers. Instead of writing a complex number as \(a + bi\) where \(a\) is the real part and \(b\) is the imaginary part, we write it in the form \(r e^{i\theta}\). Here, \(r\) is the magnitude of the complex number, calculated as \(\sqrt{a^2 + b^2}\), and \(\theta\) is the argument, the angle formed by the complex number on the complex plane, usually found using \(\tan^{-1}\left(\frac{b}{a}\right)\).
  • In our example, \(-125i\), \(a = 0\) and \(b = -125\), which means \(r = 125\) and \(\theta = -\frac{\pi}{2}\) because the direction of \(-i\) points directly downward on the complex plane, aligning with an angle of \(-90^\circ\) or \(-\frac{\pi}{2}\) radians.
  • Polar representation makes it simpler to perform operations like multiplication, division, and root extraction because the angles can be directly added or subtracted, and magnitudes are multiplied or divided.
De Moivre's Theorem
De Moivre's Theorem provides a powerful way to calculate powers and roots of complex numbers when they are in polar form. The theorem states that for any complex number \(z = r e^{i\theta}\) and integer \(n\), the formula to compute \(z^n\) is \(r^n e^{in\theta}\).
This means when raised to a power, the magnitude of the complex number is raised to that power, and its angle is multiplied by that same number. Similarly, when finding the \(n\)-th roots, the magnitude is taken as the \(n\)-th root, and the angle is divided by \(n\).
  • Applying De Moivre's Theorem to our problem: we look for the cube roots of \(125 e^{-\frac{\pi}{2}i}\).
  • The cube root means \(n = 3\), so we take \(125^{1/3} = 5\) for the magnitude and the argument \(-\frac{\pi}{2}\) divided by \(3\) gives \(-\frac{\pi}{6}\).
  • This is represented as \(z = 5 e^{i\left(-\frac{\pi}{6} + \frac{2\pi k}{3}\right)}\), where \(k\) can take the values of \(0, 1, 2\).
Cube Roots
Finding the \(n\)-th roots of a complex number, especially the cube roots, requires us to solve for multiple solutions, as each complex number will have \(n\) distinct roots. By using the formula derived from De Moivre鈥檚 Theorem, we can determine each individual root.
  • For our exercise, we calculate each root for \(k = 0, 1, 2\). This involves adjusting the angle \(\theta\): by adding \(\frac{2\pi}{3}\) times the integer steps given by \(k\).
  • For \(k = 0\), we had \(z_0 = 5 e^{-\frac{\pi}{6}i}\). Calculating it gives \(z_0 = 5(\cos(-\frac{\pi}{6}) + i\sin(-\frac{\pi}{6}))\).
  • For \(k = 1\), we shift the angle further by \(\frac{2\pi}{3}\). This gives \(z_1 = 5 e^{i\frac{\pi}{2}}\).
  • Lastly, for \(k = 2\), we do another angular step of \(\frac{2\pi}{3}\) resulting in \(z_2 = 5 e^{i\frac{7\pi}{6}}\).
Thus, the different values of \(k\) in the formula help us compute all the cube roots of the complex number.

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