/*! 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 81 In Exercises 79-82, explain the ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In Exercises 79-82, explain the mistake that is made. Find \((\sqrt{2}+i \sqrt{2})^{6}\) Solution: Raise each term to the sixth power. \((\sqrt{2})^{6}+i^{6}(\sqrt{2})^{6}\) Simplify. Let \(i^{6}=i^{4} \cdot i^{2}=-1\). \(8+8 i^{6}\) Let \(i^{6}=i^{4} \cdot i^{2}=-1 . \quad 8-8=0\) This is incorrect. What mistake was made?

Short Answer

Expert verified
The mistake was in separating terms incorrectly. The correct answer is \(-64i\).

Step by step solution

01

Identify the Expression

The problem asks us to find \((\sqrt{2} + i \sqrt{2})^6\). We are given an incorrect method of solving it, which simplifies the expression by raising each term individually to the sixth power.
02

Recognize Incorrect Method

The provided solution incorrectly separates and raises each component of the binomial \((\sqrt{2} + i \sqrt{2})\) to the sixth power. The expression was simplified to \((\sqrt{2})^6 + (i\sqrt{2})^6\). This approach is incorrect because it does not adhere to the binomial theorem or correctly handle complex numbers.
03

Understand the Correct Approach

The correct approach should treat \((\sqrt{2} + i \sqrt{2})\) as a whole and apply the binomial theorem or convert it to polar coordinates and use De Moivre's Theorem. The expression represents a complex number and needs to be handled according to the properties of complex numbers.
04

Applying the Correct Method

Convert \(\sqrt{2} + i \sqrt{2}\) to its polar form. Let's find the modulus: \(r = \sqrt{(\sqrt{2})^2 + (\sqrt{2})^2} = \sqrt{4} = 2\). The angle \(\theta\) is \(\frac{\pi}{4}\). According to De Moivre’s Theorem, \((r(cos \theta + i \sin \theta))^n = r^n (cos(n\theta) + i \sin(n\theta))\).
05

Calculate Using De Moivre’s Theorem

Using \(r = 2\), \(\theta = \frac{\pi}{4}\), and \(n = 6\), we have:\(2^6 (cos(6 \times \frac{\pi}{4}) + i \sin(6 \times \frac{\pi}{4})) = 64 (cos(\frac{3\pi}{2}) + i \sin(\frac{3\pi}{2}))\). Evaluating the cosine and sine, we find:\(cos(\frac{3\pi}{2}) = 0\) and \(\sin(\frac{3\pi}{2}) = -1\). Therefore, the result is \(64(0 - i) = -64i\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

De Moivre's Theorem
Understanding how to compute powers of complex numbers is crucial, and De Moivre's Theorem is a key tool for this. The theorem states that for any complex number expressed in polar form, \[ (r(\cos \theta + i \sin \theta))^n = r^n (\cos(n\theta) + i \sin(n\theta)) \]where \(r\) is the modulus, \(\theta\) is the argument, and \(n\) is any integer. This theorem simplifies the calculation of powers of complex numbers significantly.
De Moivre’s Theorem is particularly useful for expressions like \((\sqrt{2} + i \sqrt{2})^6\), where converting the complex number to its polar form can make calculations easier.
  • Calculate the modulus \(r\).
  • Find the argument \(\theta\).
  • Apply De Moivre’s Theorem to find the power.
By transforming a complex number to its polar form, you can effortlessly compute its powers using De Moivre's Theorem – an essential method for simplifying complex computations.
Polar Coordinates
Polar coordinates offer a fascinating way to represent complex numbers, simplifying the process of working with their powers and roots. In this system, a complex number is described in terms of its distance from the origin, known as the modulus \(r\), and the angle it forms with the positive x-axis, called the argument \(\theta\).
For the number \(\sqrt{2} + i\sqrt{2}\), its polar form is:
  • Modulus \(r\): Calculated as \(\sqrt{(\sqrt{2})^2 + (\sqrt{2})^2} = 2\).
  • Argument \(\theta\): Using trigonometry, \(\tan^{-1}\left(\frac{\sqrt{2}}{\sqrt{2}}\right) = \frac{\pi}{4}\).
This polar representation allows for easier calculations, especially using De Moivre's Theorem to find powers. Converting between rectangular and polar forms gives you a comprehensive understanding of complex numbers.
Binomial Theorem
The Binomial Theorem is a powerful mathematical tool used to expand expressions raised to a power. However, it's essential not to misapply it to complex binomials as was incorrectly done in the original exercise.
When we expand \((a + b)^n\) using the Binomial Theorem, it involves a sum of terms:\[(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\]This is not directly applicable to \((\sqrt{2} + i\sqrt{2})^6\) without converting the expression into a compatible form. It's crucial to remember that the theorem is suitable only when both terms are real numbers. For complex numbers, prefer De Moivre’s Theorem by converting them to their polar representation.
Powers of Complex Numbers
Calculating powers of complex numbers is made simpler through a combination of their polar form and fundamental theorems like De Moivre’s.
Once a complex number is in polar form, \[ (r\text{cis} \theta)^n \text{ transforms to } r^n \text{cis}(n\theta) \]where \(\text{cis} \theta\) stands for \(\cos \theta + i \sin \theta\). This powerful approach drastically reduces computation work for high powers of complex numbers.
  • High Powers: Suitable for higher powers as it avoids complex multiplication of binomial expansions.
  • Roots and Multiplication: Easily extendable to root calculations and multiplication, owing to straightforward trigonometric manipulations.
Use this method when you need to handle complex multiplication efficiently, or when working with rotational properties of complex numbers, especially in advanced algebra and geometry contexts. Understanding this process empowers you to tackle complex number problems with confidence.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Find the polar equation that is equivalent to a horizontal line, \(y=b\).

For Exercises 71 and 72, refer to the following: Spirals are seen in nature-for example, in the swirl of a pine cone. They are also used in machinery to convert motions. An Archimedes spiral has the general equation \(r=a \theta\). A more general form for the equation of a spiral is \(r=a \theta^{1 / n}\), where \(n\) is a constant that determines how tightly the spiral is wrapped. Archimedes Spiral. Compare the Archimedes spiral \(r=\theta\) with the spiral \(r=\theta^{4 / 3}\) by graphing both on the same polar graph.

For Exercises 37-46, recall that the flight of a projectile can be modeled with the parametric equations $$ x=\left(v_{0} \cos \theta\right) t \quad y=-16 t^{2}+\left(v_{0} \sin \theta\right) t+h $$ where \(t\) is in seconds, \(v_{0}\) is the initial velocity in feet per second, \(\theta\) is the initial angle with the horizontal, and \(h\) is the initial height above ground, where \(x\) and \(y\) are in feet. Bullet Fired. A gun is fired from the ground at an angle of \(60^{\circ}\), and the bullet has an initial speed of 700 feet per second. How high does the bullet go? What is the horizontal (ground) distance between where the gun was fired and where the bullet hit the ground?

In Exercises 21-36, each set of parametric equations defines a plane curve. Find an equation in rectangular form that also corresponds to the plane curve. $$ x=3 t, y=t^{2}-1 $$

In Exercises 45-68, graph each equation. In Exercises 63-68, convert the equation from polar to rectangular form first and identify the resulting equation as a line, parabola, or circle. $$ r^{2} \cos ^{2} \theta-r \sin \theta=-2 $$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.