1. ## Roots of unity

z^2-2wz-1=0, where z is a complex number and w the third root of unity. I've tried substituting w=e^(i2pi/3)

z^2-2(e^(i2pi/3))z-1=0
z^2+2*[0.5-(i*root(3)/2)]z-1=0

Using the quadratic equation gives me a square root of i, which I've never encountered before, any help is greatly appreciated, thanks in advance .

2. I havn't checked your working so far but to calculate $\displaystyle \sqrt{i}$

note that

$\displaystyle \sqrt{i}=\sqrt{e^{i\frac{\pi}{2}+2ki\pi}}=e^{i\fra c{\pi}{4}+ki\pi}=\pm \left ( \frac{1}{\sqrt 2}+i\frac{1}{\sqrt 2} \right )$

where k=0,1

3. Originally Posted by SuumEorum
z^2-2wz-1=0, where z is a complex number and w the third root of unity. I've tried substituting w=e^(i2pi/3)

z^2-2(e^(i2pi/3))z-1=0
z^2+2*[0.5-(i*root(3)/2)]z-1=0

Using the quadratic equation gives me a square root of i, which I've never encountered before, any help is greatly appreciated, thanks in advance .
From the quadratic formula: $\displaystyle z = w \pm \sqrt{w^2 + 1}$.

If $\displaystyle w = 1$ the solution is simple to express.

Otherwise note that

$\displaystyle w^3 = 1 \Rightarrow w^3 - 1 = 0$

$\displaystyle \Rightarrow (w - 1)(w^2 + w + 1) = 0 \Rightarrow w^2 + w + 1 = 0$

$\displaystyle \Rightarrow w^2 + 1 = -w$

and so $\displaystyle z = w \pm \sqrt{-w}$

etc.