Suppose w^3 = 1 and k is a positive integer
i) Find two possible values of 1 + w^k + w^2k
I have no idea where to begin? Please give me some hints.
Suppose k is a multiple of 3. Then 2k is also a multiple of 3. So 1 + w^k + w^(2k) = 3.
Suppose k gives remainder 1 when divided by 3. Then 2k gives remainder 2 when divided by 3.
Suppose k gives remainder 2 when divided by 3. Then 2k gives remainder 1 when divided by 3.
In either of the two latter cases you get 1 + w + w^2.
$\displaystyle w^3 = 1$
$\displaystyle w^3 - 1 = 0$
$\displaystyle w^3 - 1^3 = 0$
$\displaystyle (w - 1)(w^2 + w + 1) = 0$
So $\displaystyle w - 1 = 0$ or $\displaystyle w^2 + w + 1 = 0$
$\displaystyle w = 1$ or
$\displaystyle w^2 + w + \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 + 1 = 0$
$\displaystyle \left(w + \frac{1}{2}\right)^2 + \frac{3}{4} = 0$
$\displaystyle \left(w + \frac{1}{2}\right)^2 = -\frac{3}{4}$
$\displaystyle w + \frac{1}{2} = \pm\frac{\sqrt{3}i}{2}$
$\displaystyle w = \frac{-1 \pm \sqrt{3}i}{2}$.
So the three solutions are
$\displaystyle w = 1, w = \frac{-1 - \sqrt{3}i}{2}, w = \frac{-1 + \sqrt{3}i}{2}$.
Thanks for quick reply. Just got a question with the concept of this. If k gives a remainder of 4, that is equivalent to k giving a remainder of 1 right? Also, how was it acquired that if k gives remainder of 1, then 2k gives a remainder of 3 and vice versa.
Thanks really much.
k can't "give a remainder of 4" when divided by 3. The "remainder" when dividing by the integer n must be less than n.
If "k, divided by 3, gives a remainder of 1" we can write k= 3n+ 1 for some integer n. Then 2k= 2(3n+1)= 6n+ 2= 3(2n)+ 2.If you get a "remainder of 4" then you have divided incorrectly- increase your divisor by 1 and you will get a remainder of 1. Also, how was it acquired that if k gives remainder of 1, then 2k gives a remainder of 3 and vice versa.
Thanks really much.
If "2, divided by 3, gives a remaider of 2" we can write k= 3m+ 2 for some integer m. Then 2k= 2(3m+2)= 6m+ 4= 6m+ 3+ 1= 3(6m+1)+ 1.