Complex constant from single root.

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SUMMARY

The discussion focuses on determining the complex constant c such that v = √3 - j is a root of the equation z6 - c = 0. It is established that there are three distinct roots, each having a conjugate pair, leading to the conclusion that two of the six roots are √3 - j and √3 + j. The relationship z6 = c is confirmed, and the next steps involve expressing v in exponential form using the polar representation r(cosθ + jsinθ) to solve for c.

PREREQUISITES
  • Complex number theory
  • Polar and exponential forms of complex numbers
  • Roots of unity
  • Basic algebraic manipulation
NEXT STEPS
  • Explore the concept of roots of unity in complex analysis
  • Learn how to convert complex numbers to polar form
  • Study the properties of complex conjugates
  • Investigate the use of De Moivre's Theorem for finding roots
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Students studying complex analysis, mathematicians working with polynomial equations, and anyone interested in understanding the properties of complex roots and their applications.

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Homework Statement


Do not use a calculator for this problem. Express your answers using square roots and/or fractional multiples of x.

Determine the complex constant c such that v is a root of: z6 - c = 0


Homework Equations


v = \sqrt{3} - j


The Attempt at a Solution



I believe the following are true:
1. There are 3 distinct roots for this equation.
2. Each of the distinct roots will have a conjugate pair to ensure there is no middle term.
3. If 2 is true, then two of the six roots are \sqrt{3} -j and \sqrt{3} +j

z6 - c = 0 --> z6 = c. Yes, this is as far as I have gotten. I'm not sure how to figure out the other 4 roots.
 
Physics news on Phys.org
c = z6
v = √3 - j
Use re = r(cosθ + jsinθ)
Write v as exponential
Solve for c
 

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