gday y'all,
what's the quickest way to do this?
also can someone explain this: solutions show k=0,1,2,3,4,5,6 but this is out of the restriction as -pi=<theta=<pi, correct results would be k=0,+/-1,+/-2,+/-3 ?
thanks in advance
View attachment 32576
First thing I would to is convert this to exponential form. It's way quicker this way. Also it doesn't matter the order as long as its consecutive, in theory you could do 22,23,24,25... you get my point - but that's inefficient. If the angle is greater than pi, you would just find the related angle and you should get the same values as you would k = 0, +/-1 and so on, it is more efficient to do that mirrored format of -3 to 3 since you would get the related angles already most of the times, and since its an odd integer, it would be symmetrical.
I'm not sure if you're looking for solutions but this is what I did.
z^7 = 1
z^7 = cis(pi(2k)) (Apologies purists. It hurts me too.)
z = cis(pi/7*(2k)) = e^pi(2k)/7
Do your regular substitution of values, make use of the symmetry of odd integers as there can only be 7 roots at max. so k =-3 to k=3.
I would just find the roots from -3 to 0 and then stop since thats all the roots you'll need, the other side from 0 to 3 is the same as k=-3 to 0 but its just the opposite sign so in that way you can save time.
Now you found all the roots, you would just select the smallest value root and that would be 2pi/7. You can alternatively plot the points as a circle on an argand diagram and find the solution that way if that's your fancy.
Here's how I showed that the other roots were solutions;
z^7 -1 = 0
Substituting w^k, where k is a positive integer
(w^k)^7 - 1 =0
Therefore;
(w^7)^k -1 =0
Hence w^7 is a root and we can deduce that roots before it, precisely 0 to 7 are all solutions of the equation as well. One root is 1 so the total roots would be;
1, w, w^2,w^3,w^4,w^5,w^6
as req.
Not claiming this is the fastest way but this usually takes me 2-3 minutes or 1-2 if I really want to be fast, it becomes formulaic and repetitive after a while when you do these type of questions so getting it down in time shouldn't be an issue. If anyone has a faster way or even another form of solution then it would be appreciated
