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

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SUMMARY

The discussion focuses on solving the equation z3 = -8 in the context of complex numbers. Participants suggest converting -8 into polar coordinates, represented as -8 = 8(cos 0 + i sin 0), and finding the cube roots using the formula r = (-8)^(1/3). The conversation highlights the use of D'Moivre's theorem and the cube roots of unity as essential concepts for deriving the three roots of the equation.

PREREQUISITES
  • Understanding of complex numbers and their polar representation
  • Familiarity with D'Moivre's theorem
  • Knowledge of cube roots of unity
  • Basic polynomial equations and the factor theorem
NEXT STEPS
  • Study the application of D'Moivre's theorem in complex number calculations
  • Learn how to derive cube roots of unity and their properties
  • Explore polar coordinates and their use in complex number analysis
  • Investigate the factor theorem and its application in solving polynomial equations
USEFUL FOR

Students studying complex numbers, mathematics educators, and anyone interested in solving polynomial equations in the complex plane.

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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