How Do You Solve Complex Number Equations?

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SUMMARY

The discussion focuses on solving the complex number equation z^3 + 8 = 0. The first attempt incorrectly identifies one root as z = 2i, while the second attempt correctly identifies z = -2 as a root but fails to find the remaining roots. The correct approach involves using polynomial division to factor z^3 + 8 by (z + 2), leading to a quadratic equation that can be solved for the other two roots. This method ensures all three roots of the cubic equation are identified.

PREREQUISITES
  • Understanding of complex numbers and their properties
  • Familiarity with polynomial equations and their roots
  • Knowledge of polynomial division techniques
  • Basic understanding of De Moivre's Theorem (for future reference)
NEXT STEPS
  • Study polynomial division methods for finding roots of cubic equations
  • Learn about De Moivre's Theorem and its application in finding complex roots
  • Explore the polar form of complex numbers and its significance
  • Practice solving higher-degree polynomial equations and identifying all roots
USEFUL FOR

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

missmerisha
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I hope, I've posted this question in the right section.

Homework Statement


Solve the fooling equation over C

z^3+ 8 = 0


The Attempt at a Solution



First Attempt
z^3 = -8
cube root (2 ^3) = cube root (8 i^2 )
z = 2i


Second Attempt
z^3 = -8
z ^3 = -2 ^3
so, z = -2
 
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Or you could convert -8 into polar form, then using De Moivre's theorem get all three cube roots.
 
We're learning Polar Form next year and I have never heard of De Moivre's Theorem.

So, is my second attempt incorrect?
 
Your first attempt is incorrect; (2i)^3=-8i\neq -8

Your second attempt is not incorrect but it is incomplete: z^3+8=0 is a 3rd degree polynomial equation; so it must have three roots. You have correctly found one root z=-2, but you still need to find the other two.

One method is to divide your polynomial z^3+8 by z+2 (Since z=-2 is a root, you know (z+2) must be a factor of the polynomial) which will leave you with a quadratic that you can solve to find your other two roots.
 
thanks
I've got it now.
 

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