Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Problem 131

\([\\{1+\cos (\pi / 12)+i \sin (\pi / 12)\\} /\\{1+\cos (\pi / 12)\) \(-i \sin (\pi / 12)\\}]^{36}=\) (a) \(-1\) (b) 1 (c) 0 (d) \((1 / 2)\)

Short Answer

Expert verified
The given complex expression simplifies to \(1\), and when raised to the power of \(36\), it remains \(1\), which corresponds to option (b).

Step by step solution

01

Simplify the Expression

To start, let's simplify the given expression: \[ \frac{1+\cos(\frac{\pi}{12})+i\sin(\frac{\pi}{12})}{1+\cos(\frac{\pi}{12})-i\sin(\frac{\pi}{12})} \] One way to rationalize this complex number is to multiply the numerator and the denominator by the conjugate of the denominator: \[ \frac{(1+\cos(\frac{\pi}{12})+i\sin(\frac{\pi}{12})) (1+\cos(\frac{\pi}{12})+i\sin(\frac{\pi}{12}))}{(1+\cos(\frac{\pi}{12})-i\sin(\frac{\pi}{12}))(1+\cos(\frac{\pi}{12})+i\sin(\frac{\pi}{12}))} \]
02

Multiply and Simplify

Next, we multiply and simplify the expression: \[ \frac{1+2\cos(\frac{\pi}{12})+\cos^2(\frac{\pi}{12}) - \sin^2(\frac{\pi}{12})}{1+2\cos(\frac{\pi}{12})+\cos^2(\frac{\pi}{12})+\sin^2(\frac{\pi}{12})} \] Now, note that \(\cos^2(\frac{\pi}{12})+\sin^2(\frac{\pi}{12})=1\) . So, we can replace this sum with 1: \[ \frac{1+2\cos(\frac{\pi}{12})+(1- \sin^2(\frac{\pi}{12}))}{1+2\cos(\frac{\pi}{12})+1} \] Further simplifying, we obtain: \[ \frac{2(1+\cos(\frac{\pi}{12}))}{2+2\cos(\frac{\pi}{12})} \]
03

Cancel terms

Now, cancel the common terms in the numerator and the denominator: \[ \frac{1+\cos(\frac{\pi}{12})}{1+\cos(\frac{\pi}{12})} = 1 \]
04

Raise to the 36th power

Now, we need to raise the result we found in the previous step to the power of 36: \[ (1)^{36} = 1 \] The result matches option (b) which is 1.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

If \(z\) is a complex number satisfying \(|z-\operatorname{iRe}(z)|=|z-\operatorname{Im}(z)|\) then \(\mathrm{z}\) lies on (a) \(\mathrm{y}=\mathrm{x}-1\) (b) \(\mathrm{y}=\pm \mathrm{x}\) (c) \(\mathrm{y}=\mathrm{x}+1\) (d) \(\mathrm{y}=-\mathrm{x}+1\)

The complex numbers \(\mathrm{z}_{1}, \mathrm{z}_{2}\) and \(\mathrm{z}_{3}\) satisfying \(\left[\left(z_{1}-z_{3}\right) /\left(z_{2}-z_{3}\right)\right]=[(1-i \sqrt{3}) / 2]\) are the vertices of a triangle which is (a) of area zero (b) right angled isosceles (c) equilateral (d) obtuse-angle isosceles

The inequality \(|z-4|<|z-2|\) represent the region given by (a) \(\operatorname{Re}(z)>0\) (b) \(\operatorname{Re}(z)<0\) (c) \(\operatorname{Re}(z)>2\) (d) \(\operatorname{Re}(z)>3\)

If \(z=x+y\) i and \(|3 z|=|z-4|\) then \(x^{2}+y^{2}+x=\) (a) 1 (b) \(-1\) (c) 2 (d) \(-2\)

If \(\mathrm{x}=\mathrm{a}+\mathrm{b}, \mathrm{y}=\mathrm{a} \alpha+\mathrm{b} \beta\) and \(\mathrm{z}=\mathrm{a} \beta+\mathrm{b} \alpha\), Where \(\alpha, \beta \neq 1\) are cube roots of unity, then \(\mathrm{xyz}=\) (a) \(2\left(\mathrm{a}^{3}+\mathrm{b}^{3}\right)\) (b) \(2\left(a^{3}-b^{3}\right)\) (c) \(\mathrm{a}^{3}+\mathrm{b}^{3}\) (d) \(a^{3}-b^{3}\)
See all solutions

Recommended explanations on English Textbooks

Language Acquisition

Read Explanation

Discourse

Read Explanation

Single Paragraph Essay

Read Explanation

Pragmatics

Read Explanation

Rhetorical Analysis Essay

Read Explanation

Phonetics

Read Explanation
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.

Sign-up for free