Complex numbers, i don't kno if it's before or after calculus sry

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Homework Help Overview

The problem involves finding the three roots of the equation z3 = 1, specifically focusing on the non-real roots expressed in the form of e within the interval -π < θ ≤ π. The context includes complex numbers and potentially relates to roots of unity.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the identification of roots, with one noting that one root is 1 and the others form a conjugate pair. There are references to De Moivre's theorem and the polar form of complex numbers. Some participants suggest exploring the roots of unity and polynomial factorization as methods to find the remaining roots.

Discussion Status

The discussion is ongoing, with participants offering various approaches and insights. Some guidance has been provided regarding the use of De Moivre's theorem and polynomial factorization, but there is no explicit consensus on the best method to proceed.

Contextual Notes

One participant mentions a lack of familiarity with De Moivre's theorem and indicates that the problem may not be covered in their current syllabus, suggesting a potential gap in knowledge relevant to the discussion.

Kushal
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Homework Statement



find the three roots of z[tex]^{3}[/tex] = 1, giving non real roots in the form of e[tex]^{i\theta}[/tex], where -[tex]\pi[/tex]<[tex]\theta[/tex][tex]\leq[/tex][tex]\pi[/tex]

Homework Equations



z = re[tex]^{i\theta}[/tex]

The Attempt at a Solution



i kno that one of the roots is 1, an the other two form a conjugate pair. i can't find them.
 
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google "roots of unity" and read about how you can locate them. They are spaced evenly on the unit circle.
 
and the question continues, i can't find that too.

given that [tex]\omega[/tex] is one of the non real roots, show that 1 + [tex]\omega[/tex] + [tex]\omega^{2}[/tex] = 0
 
thanks futurebird
 
In general from de moivre’s theorem, when you consider a complex number in polar form you get

z^(1/n) = r(cosx + isinx)^1/n = r^(1/n)(cos([x+2kpi]/n) + I*sin([x+2kpi]/n)) for 0<k<n-1 where k is an integer
 
the thing is that I'm not familiar with de moivre's theorem. i guess we did only basic complex numbers at school. i came across this question on a past paper from another examination board. it's prolly not in our syllabus.

thnks
 
Kushal said:
z = re[tex]^{i\theta}[/tex]
I assume you know what this equation means, i.e. that this is a possible way to express a complex number (and that r>=0). Do you also know how to multiply two complex numbers?
[tex]z_1 \cdot z_2 = r_1 e^{i\theta_1} \cdot r_2 e^{i\theta_2} = r_1 r_2 e^{i(\theta_1 + \theta_2)}[/tex], where [tex]e^{i2\pi} = 1[/tex] might be needed, in case you didn't already know that one.

i know that one of the roots is 1, an the other two form a conjugate pair. i can't find them.
You are looking for z such that z*z*z = 1. Write that out in the form given and try to find solutions for z.
 
Well if you knew that z=1 was a solution to the equation [itex]z^3=1 => z^3-1=0[/itex] would that not mean that [itex](z-1)[/itex] is a factor of [itex]z^3-1=0[/itex] ?

then you can just divide it out and you'll get [itex](z-1)(z^2+bz+1)=0[/itex] and then find the value of b...and then you use the quad. eq'n formula on the 2nd factor and get your roots...then put those roots into the form [itex]re^{i\theta}[/itex]

but it would be easier to use the roots of unity method
 

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