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


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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
	$z^3 =1 {z}={e^(i((2π)/3)k) }_(0≤k<3) Σ_(k=0) ^2 e^(i((2π)/3)k) =Σ_(k=0) ^2 (e^(i((2π)/3)) )^k =(((e^(i((2π)/3)) )^3 −1)/(e^(i((2π)/3)) −1))=((e^(i2π) −1)/(e^(i((2π)/3)) −1))= =((1−1)/(e^(i((2π)/3)) −1))=0$
	$z^{3} = 1$ ${z} = {e^{i \frac{2 π}{3} k}}_{0 ⩽ k < 3}$ $\underset{k = 0}{\sum^{2}} e^{i \frac{2 π}{3} k} = \underset{k = 0}{\sum^{2}} {(e^{i \frac{2 π}{3}})}^{k} = \frac{{(e^{i \frac{2 π}{3}})}^{3} - 1}{e^{i \frac{2 π}{3}} - 1} = \frac{e^{i 2 π} - 1}{e^{i \frac{2 π}{3}} - 1} =$ $= \frac{1 - 1}{e^{i \frac{2 π}{3}} - 1} = 0$
	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

	$z^{3} = 1 \Rightarrow z = e^{i \frac{2 k π}{3}}; k \in [∣ 0; 2 ∣]$ $\to k = 0 \Rightarrow z = 1$ $\to k = 1 \Rightarrow z = e^{i \frac{2 π}{3}}$ $\to k = 2 \Rightarrow z = e^{i \frac{4 π}{3}} = e^{- i \frac{2 π}{3}}$ $l e t j = e^{i \frac{2 π}{3}}$ $1 + j + j^{2} = \frac{1 - j^{3}}{1 - j} = \frac{{(e^{i \frac{2 π}{3}})}^{3} - 1}{e^{i \frac{2 π}{3}} - 1} = 0 . . .$ $- - - - - - - - - - - - -$ $j = e^{i \frac{2 π}{3}} = - \frac{1}{2} + \frac{\sqrt{3}}{2} i a n d$ $j^{2} = e^{i \frac{4 π}{3}} = e^{- i \frac{2 π}{3}} = - \frac{1}{2} - \frac{\sqrt{3}}{2} i . .$ $1 + j + j^{2} = 1 - \frac{1}{2} + \frac{\sqrt{3}}{2} i - \frac{1}{2} - \frac{\sqrt{3}}{2} i = 1 - 1 = 0$ $\Rightarrow 1 + j + j^{2} = 0 . . .$
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