# Thread: In the complex plane

1. ## In the complex plane

Question 1)
In the complex plane , there are three cube roots of one. Let zeta be the cube root of one which has positive imaginary part. Show that zeta is a quadratic integer in Q[sqrt(-3)] by writing it in the form (a+bsqrt(-3))/2 , where a and b are rational integers and a and b are either both even or both odd. Then write it in the form m+n*((1+sqrt(-3))/2), where m and n are rational integers.

I have totally no idea what the questions about.
Could you please give me detail explaination if you can solve them? Thank you very much.

Note : I have deleted Question 2 . Since no one reply for it and I have just solved it.

2. Originally Posted by beta12
Question 1)
In the complex plane , there are three cube roots of one. Let zeta be the cube root of one which has positive imaginary part. Show that zeta is a quadratic integer in Q[sqrt(-3)] by writing it in the form (a+bsqrt(-3))/2 , where a and b are rational integers and a and b are either both even or both odd. Then write it in the form m+n*((1+sqrt(-3))/2), where m and n are rational integers.
The cube root of unity that we want is: $\displaystyle z=-1/2 + \sqrt{3}\ i/2=\frac{-1+\sqrt{-3}}{2}$.

Which is in the first form required with $\displaystyle a=-1,\ b=1$, and the second form with
$\displaystyle m=-1,\ n=1$.

That the root of unity that we require is that given we need to write:

$\displaystyle 1=e^{2n \pi i},\ \ n=0,\ \pm 1,\ ...$

Then the cube roots of unity are:

$\displaystyle z=e^{2n \pi i/3},\ \ n=0,\ \pm 1,\ ...$

But only three of these are distinct and are given by:

$\displaystyle z=e^{2n \pi i/3},\ \ n=0,\ \pm 1$

Writing these out in the form $\displaystyle a+bi$ shows that the one given
above is the one required.

RonL

3. Hi Captainblack,

Thank you for your reply. I will go through it in detail. If I don't understand , I will come back to you. Thank you very much again.