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

[Kushal]

Homework Statement

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

Homework Equations

z = re^{i\theta}

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.

 

[futurebird]

google "roots of unity" and read about how you can locate them. They are spaced evenly on the unit circle.

 

[Kushal]

and the question continues, i can't find that too.

given that \omega is one of the non real roots, show that 1 + \omega + \omega^{2} = 0

 

[Kushal]

thanks futurebird

 

[JonF]

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

 

[Kushal]

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

 

[Timo]

z = re^{i\theta}

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?
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)}, where e^{i2\pi} = 1 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.

 

[rock.freak667]

Well if you knew that z=1 was a solution to the equation z^3=1 =&gt; z^3-1=0 would that not mean that (z-1) is a factor of z^3-1=0 ?

then you can just divide it out and you'll get (z-1)(z^2+bz+1)=0 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 re^{i\theta}

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