Chapter 8: Problem 57
Convert all complex numbers to trigonometric form and then simplify each expression. Write all answers in standard form. $$ \frac{(1+i)^{4}(2 i)^{2}}{-2+2 i} $$
Short Answer
Expert verified
The expression simplifies to \(-4 - 4i\).
Step by step solution
01
Convert Complex Numbers to Polar Form
Start with identifying the modulus and angle for each complex number. For \((1+i)\), the modulus \( r = \sqrt{1^2 + 1^2} = \sqrt{2} \) and the angle \( \theta = \tan^{-1}(\frac{1}{1}) = \frac{\pi}{4} \). For \(2i\), the modulus \(r = 2\) and angle \(\frac{\pi}{2}\), since it lies on the imaginary axis. For \(-2 + 2i\), the modulus \(r = \sqrt{(-2)^2 + 2^2} = 2\sqrt{2}\) and angle \(\tan^{-1}(\frac{2}{-2}) = \frac{3\pi}{4}\).
02
Express in Polar Form and Simplify
Convert each complex number into polar form using Euler's formula: \((1+i)^4 = (\sqrt{2})^4 (e^{i\frac{\pi}{4}})^4 = 4e^{i\pi}\), \((2i)^2 = 2^2 e^{i\frac{\pi}{2}\cdot 2} = 4e^{i\pi}\), \(-2+2i = 2\sqrt{2} e^{i\frac{3\pi}{4}}\). Now substitute into the expression: \[\frac{(1+i)^{4}(2 i)^{2}}{-2+2 i} = \frac{4e^{i\pi} \cdot 4e^{i\pi}}{2\sqrt{2} e^{i\frac{3\pi}{4}}}\].
03
Simplify Using Polar Form Properties
Simplify the numerator: \(4e^{i\pi} \cdot 4e^{i\pi} = 16e^{i(2\pi)} = 16(1)\) since \(e^{i2\pi} = 1\). The denominator is \(2\sqrt{2} e^{i\frac{3\pi}{4}}\). The expression becomes: \[\frac{16}{2\sqrt{2}} e^{-i\frac{3\pi}{4}}\].
04
Simplify the Expression to Standard Form
Divide the modulus: \(\frac{16}{2\sqrt{2}} = 4\sqrt{2}\), leaving \(4\sqrt{2}e^{-i\frac{3\pi}{4}}\). Convert back to standard form: \(e^{-i\frac{3\pi}{4}} = \cos(-\frac{3\pi}{4}) + i\sin(-\frac{3\pi}{4}) = -\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}\). Thus, multiply by \(4\sqrt{2}\):\[4\sqrt{2}(-\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}) = -4 - 4i\].
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.
Complex Numbers
Complex numbers are a fascinating field of mathematics where each number can be represented as a sum of a real part and an imaginary part. The standard form for a complex number is written as:
The modulus and argument (angle) of a complex number provide an alternative representation, forming the basis of its conversion to polar form. The modulus is the distance from the origin to the point, calculated as:
- \( z = a + bi \)
The modulus and argument (angle) of a complex number provide an alternative representation, forming the basis of its conversion to polar form. The modulus is the distance from the origin to the point, calculated as:
- \( r = \sqrt{a^2 + b^2} \)
Polar Form
The polar form of a complex number provides an elegant and insightful way to analyze and work with complex numbers, particularly when multiplying or taking powers. This form presents a complex number in terms of its modulus \( r \) and argument \( \theta \):
- \( z = r (\cos \theta + i \sin \theta) \)
- This expression is compactly represented using Euler's formula as \( z = r e^{i\theta} \).
- For multiplication, simply multiply the moduli and add the arguments.
- For division, divide the moduli and subtract the arguments.
Euler's Formula
Euler's formula is a cornerstone in understanding the trigonometric form of complex numbers. It is expressed as:
A remarkable feature of Euler's formula is how it relates to the unit circle in trigonometry:
- \( e^{i\theta} = \cos \theta + i \sin \theta \)
A remarkable feature of Euler's formula is how it relates to the unit circle in trigonometry:
- \( e^{i\theta} \) represents a point on the unit circle in the complex plane.
- The real part \( \cos \theta \) and the imaginary part \( \sin \theta \) trace a circular path with radius 1 as \( \theta \) varies.