Find-the-cube-root-of-one-Hence-show-that-the-sum-of-the-root-is-equal-to-zero-

Question Number 155571 by peter frank last updated on 02/Oct/21

Find the cube root of one . Hence

show that the sum of the root is

equal to zero

Answered by JDamian last updated on 02/Oct/21

z^{3} = 1

{z} = {e^{i \frac{2 π}{3} k}}_{0 ⩽ k < 3}

\underset{k = 0}{\sum^{2}} e^{i \frac{2 π}{3} k} = \underset{k = 0}{\sum^{2}} {(e^{i \frac{2 π}{3}})}^{k} = \frac{{(e^{i \frac{2 π}{3}})}^{3} - 1}{e^{i \frac{2 π}{3}} - 1} = \frac{e^{i 2 π} - 1}{e^{i \frac{2 π}{3}} - 1} =

= \frac{1 - 1}{e^{i \frac{2 π}{3}} - 1} = 0

Commented by peter frank last updated on 02/Oct/21

great sir; thanks

Commented by puissant last updated on 02/Oct/21

N i c e s i r . . b u t i t i s r a t h e r k \in {0, 1, 2} . .

Answered by puissant last updated on 02/Oct/21

z^{3} = 1 \Rightarrow z = e^{i \frac{2 k π}{3}}; k \in [∣ 0; 2 ∣]

\to k = 0 \Rightarrow z = 1

\to k = 1 \Rightarrow z = e^{i \frac{2 π}{3}}

\to k = 2 \Rightarrow z = e^{i \frac{4 π}{3}} = e^{- i \frac{2 π}{3}}

l e t j = e^{i \frac{2 π}{3}}

1 + j + j^{2} = \frac{1 - j^{3}}{1 - j} = \frac{{(e^{i \frac{2 π}{3}})}^{3} - 1}{e^{i \frac{2 π}{3}} - 1} = 0 \dots

- - - - - - - - - - - - -

j = e^{i \frac{2 π}{3}} = - \frac{1}{2} + \frac{\sqrt{3}}{2} i a n d

j^{2} = e^{i \frac{4 π}{3}} = e^{- i \frac{2 π}{3}} = - \frac{1}{2} - \frac{\sqrt{3}}{2} i . .

1 + j + j^{2} = 1 - \frac{1}{2} + \frac{\sqrt{3}}{2} i - \frac{1}{2} - \frac{\sqrt{3}}{2} i = 1 - 1 = 0

\Rightarrow 1 + j + j^{2} = 0 \dots

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