Q. 26 Find all the complex cube roots ... [FREE SOLUTION]

91影视

Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 9: Q. 26 (page 630)

Find all the complex cube roots of 27.

Short Answer

Expert verified

$z_{0} = 3 z_{1} = 3 [\cos 120^{鈭�} + i \sin 120^{鈭�}] z_{2} = 3 [\cos 240^{鈭�} + i \sin 240^{鈭�}]$

Step by step solution

Step 1. Write the formula for the complex root of a number in form w=r(cosθo+isinθo)

$z_{k} = \sqrt[n]{r} [\cos (\frac{胃_{0}}{n} + \frac{2 k 蟺}{n}) + i \sin (\frac{胃_{0}}{n} + \frac{2 k 蟺}{n})]$

Step 2. Find r and θ

$r = \sqrt{a^{2} + b^{2}} = \sqrt{27^{2} + 0^{2}} = 27 胃 = \tan^{- 1} (\frac{y}{x}) = \tan^{- 1} (\frac{0}{27}) = \tan^{- 1} (0) = 0^{o}$

Step 3. Write the polar form of the given equation

Polar form will be:

$27 = 27 (\cos 0^{o} + i \sin 0^{o})$

Step 4. Find the complex cube roots

$z_{k} = \sqrt[3]{27} [c o s (\frac{0^{鈭�}}{3} + \frac{360^{鈭�} k}{3}) + i s i n (\frac{0^{鈭�}}{3} + \frac{360^{鈭�} k}{3})] = 3 [c o s (0^{鈭�} + 120^{鈭�} k) + isin (0^{鈭�} + 120^{鈭�} k)]$

For k = 0

$z_{0} = 3 [\cos (0^{鈭�} + 120^{鈭�} (0)) + i \sin (0^{鈭�} + 120^{鈭�} (0))] = 3 [\cos 0^{鈭�} + i \sin 0^{鈭�}] = 3$

For k = 1

$z_{1} = 3 [\cos (0^{鈭�} + 120^{鈭�} (1)) + i \sin (0^{鈭�} + 120^{鈭�} (1))] = 3 [\cos 120^{鈭�} + i \sin 120^{鈭�}]$

For k = 2

$z_{2} = 3 [\cos (0^{鈭�} + 120^{鈭�} (2)) + i \sin (0^{鈭�} + 120^{鈭�} (2))] = 3 [\cos 240^{鈭�} + i \sin 240^{鈭�}]$

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91影视

Find all the complex cube roots of 27.

Short Answer

Step by step solution

Step 1. Write the formula for the complex root of a number in form w=r(cosθo+isinθo)

Step 2. Find r and θ

Step 3. Write the polar form of the given equation

Step 4. Find the complex cube roots

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