Q. 63 聽Show that each complex nth roo... [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. 63 (page 592)

Show that each complex nth root of a nonzero complex number w has the same magnitude.

Short Answer

Expert verified

$z_{n} = r ({肠辞蝉胃}_{慰} + {颈蝉颈苍胃}_{慰}) = w$

Step by step solution

Step 1. Considering a complex number w

Assume $w = r (\cos 胃_{鈭�} + i \sin 胃_{鈭�})$ is any non zero complex number.

The m distinct roots of are ;

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

n=0,1,2,3........(m-1)

m is greater than or equal to 2.

Step 2. Calculations

Complex $m^{t h}$ of non zero complex number is

${z_{n}}^{m} = {\sqrt[m]{r} [(\cos (\frac{胃_{慰}}{m} + \frac{2 n 蟺}{m}) + i \sin (\frac{胃_{慰}}{m} + \frac{2 n 蟺}{m}))]}^{m} = {(\sqrt[m]{r})}^{m} (\cos (m) (\frac{胃_{慰}}{m} + \frac{2 n 蟺}{m}) + i \sin (m) (\frac{胃_{慰}}{m} + \frac{2 n 蟺}{m}))$ $= r [\cos (胃_{慰} + 2 n 蟺) + isin (胃_{慰} + 2 苍蟺)]$

Step 3. Using trigonometric formula

$= r [\cos (胃_{慰} + 2 n 蟺) + isin (胃_{慰} + 2 苍蟺)] \cos (胃_{慰} + 2 n 蟺) = \cos 胃_{慰} \sin (胃_{慰} + 2 苍蟺) = {蝉颈苍胃}_{慰} z_{n} = r ({肠辞蝉胃}_{慰} + {颈蝉颈苍胃}_{慰}) = w$

Hence it has proved that each complex nth root of a nonzero complex number w has the same magnitude.

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91影视

Show that each complex nth root of a nonzero complex number w has the same magnitude.

Short Answer

Step by step solution

Step 1. Considering a complex number w

Step 2. Calculations

Step 3. Using trigonometric formula

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