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I need to find the solution of $z^{3} =-1$. Let$$z=[ r,\theta ]$$ then using de Moivre's theorm$$ z^{3} =\left[ r^{3} ,3\theta \right] $$ write the number $-1$ in modulus argument from, $$[ 1,( 2n-1) \pi ]$$ the we can write$$\left[ r^{3} ,3\theta \right] =[ 1,( 2n-1) \pi ]$$ Therefore,$$ r^{3} =1 \space \text{and} \space \theta =\frac{( 2n-1) \pi }{3}$$ Then$$z_{n} =1\left( \cos\frac{( 2n-1) \pi }{3} +i\sin\frac{( 2n-1) \pi }{3}\right)$$ \begin{align} &n=0 \quad z_{0} =\frac{1}{2} -i\frac{\sqrt{3}}{2}\\ &n=1 \quad z_{1} =\frac{1}{2} +i\frac{\sqrt{3}}{2}\\ &n=2 \quad z_{1} =-1\\ \end{align}

Starlight
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RGen
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  • Your answer is correct. You can see this from: Your three solutions are clearly solutions and distinct. As this can only have three distinct solutions, these are all solutions. – Lazy Dec 14 '21 at 07:27
  • "write the number −1 in modulus argument from ..." $;-;$ Except the modulus must be a non-negative number, and the argument is not $,n \pi,$. – dxiv Dec 14 '21 at 07:28
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    @Lazy No, the answer is not correct. The first red flag is that the complex roots should be conjugates, while OP's are not. – dxiv Dec 14 '21 at 07:29
  • You have mentioned complex root of unity, but then solved z^3=-1. If you want the complex roots of unity, shouldn't it be z^3=1. – Starlight Dec 14 '21 at 07:30
  • I'm following the lecture note, It has not any examples. It noted that Complex root of unity. @Starlight I'm sorry, I don't know much about that. – RGen Dec 14 '21 at 07:33
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    @dxiv Oh yeah, I didn’t read closely enough to spot that there is a big modulus error (that’s what being tired does for you). Basically the error is that you do not want $(-1,n\pi)$ but $(-1,n2\pi)$, as $e^{it}$ is $2\pi$-periodic. So then $\theta = 2n\pi/3$. – Lazy Dec 14 '21 at 07:35
  • @dxiv, thanks sir – RGen Dec 14 '21 at 07:36
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    Note that unity means 1, not (-1). – Starlight Dec 14 '21 at 07:36
  • @Lazy now is it correct? updated – RGen Dec 14 '21 at 07:40
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    @Lazy The canonical modulus-argument representation is $,\big(1, (2n+1)\pi\big),$. – dxiv Dec 14 '21 at 07:41
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    @dxiv I think it is correct now. Thank you sir – RGen Dec 14 '21 at 07:48
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    @dxiv That is correct, but irrelevant for the construction. – Lazy Dec 14 '21 at 07:50

2 Answers2

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You can find the complex roots of unity as below. The equation you have solved is not the one that will give you complex roots of unity.

$$z^3=1 \implies z^3-1=0$$ Factor the LHS (using any of a variety of methods): $$(z-1)(z^2+z+1)=0$$ Use the zero-product property: $$z-1=0\implies z=1$$ $$z^2+z+1=0\implies z=\frac{-1\pm i\sqrt{3}}{2}$$ (where we used the quadratic formula above)

Starlight
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    Thanks for your effort, But I need to solve above equation by using complex complex number de moivre's theorem. – RGen Dec 14 '21 at 07:26
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$z=re^{i\theta}$

$z^3 = -1$

$r^3 (e^{3 i \theta})= 1\cdot e^{i\pi}$

Hence, $ r=1$ and $\theta= \frac{i(\pi + 2k\pi) }{3}$ for $k=0, 1,2$

$z= e^{\frac{i\pi}{3}} , e^{i\pi}, e^{\frac{5i\pi}{3}}$

Sourav Ghosh
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