How to Solve z^3 = -8 in Complex Numbers?

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Homework Help Overview

The discussion revolves around solving the equation z3 = -8 within the context of complex numbers. Participants explore various methods to find the complex roots of this equation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss converting -8 into polar coordinates and consider the implications of using D'Moivre's theorem. There are questions about the representation of -8 in polar form and the relevance of cube roots of unity in the solution process.

Discussion Status

The conversation is active, with participants offering different perspectives on how to approach the problem. Some suggest using polar coordinates while others mention alternative methods like the remainder and factor theorem. There is no explicit consensus on a single approach yet.

Contextual Notes

One participant expresses confusion about the problem's context, indicating a potential gap in foundational knowledge regarding complex numbers and their properties. The discussion also hints at the appropriateness of posting homework questions in the forum.

sara_87
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find all complex numbers satisfying z^3=-8
i have no idea how to solve this can someone help please?
thank u
 
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Think -8 as -8 + 0i... Then convert this into polar coordinates.
 
in polar coordinates it will be:
-8=8(cos0+isin0)
giving
-8=8
?
 
i think that if you write down z=rexp(ix) would be easier...
you find three roots on the cricle with radius r=(-8)^(1/3).

bye
marco
 
Are you saying you have been asked to find cuberoots and have never heard of D'Moivre's theorem? Why that's awful! Almost as bad posting a homework problem under the mathematics thread!
 
He actually doesn't need D'Moivre's theorem, only the three cube roots of unity.

Hi sara_87,

Have you done the complex cube roots of one in class? You know, omega, omega^2 etc.? Show some work.
 
Or just basic remainder and factor theorem...f(z)=z^3+8=0. But the other ways are faster.
 

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