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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

$z_{n} = r ({cosθ}_{ο} + {isinθ}_{ο}) = w$

Step by step solution

01

Step 1. Considering a complex number w

Assume $w = r (\cos θ_{\circ} + i \sin θ_{\circ})$ is any non zero complex number.

The m distinct roots of are ;

$z_{n} = \sqrt[m]{r} [(\cos (\frac{θ_{ο}}{m} + \frac{2 n π}{m}) + i \sin (\frac{θ_{ο}}{m} + \frac{2 n π}{m}))]$

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

m is greater than or equal to 2.

02

Step 2. Calculations

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

${z_{n}}^{m} = {\sqrt[m]{r} [(\cos (\frac{θ_{ο}}{m} + \frac{2 n π}{m}) + i \sin (\frac{θ_{ο}}{m} + \frac{2 n π}{m}))]}^{m} = {(\sqrt[m]{r})}^{m} (\cos (m) (\frac{θ_{ο}}{m} + \frac{2 n π}{m}) + i \sin (m) (\frac{θ_{ο}}{m} + \frac{2 n π}{m}))$ $= r [\cos (θ_{ο} + 2 n π) + isin (θ_{ο} + 2 nπ)]$

03

Step 3. Using trigonometric formula

$= r [\cos (θ_{ο} + 2 n π) + isin (θ_{ο} + 2 nπ)] \cos (θ_{ο} + 2 n π) = \cos θ_{ο} \sin (θ_{ο} + 2 nπ) = {sinθ}_{ο} z_{n} = r ({cosθ}_{ο} + {isinθ}_{ο}) = w$

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

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