Complex eigenvectors help
I'm currently working on a physics problem involving six identical masses arranged in a circle, attached to each other via identical springs.
My job is to find the normal modes using a symmetry matrix and its corresponding eigenvectors and eigenvalues. Since I'm working with 6 masses, I get a 6x6 symmetry matrix to fulfill the eigenvector/eigenvalue equation SA=BA, where S^6=I, which means there are six normal modes (some are complex) that I can work with.
The problem is that when I tried solving for the general form of the eigenvector B, I found that the components of B (I'll denote them b) follow this fancy equation: b^6=1. I already know that this has 2 real values and 4 complex ones, but first I need to provide a general formula that covers all the complex values.
As in, I can't just leave my answer like <-1,1, etc., etc., etc., etc.>, but I need to provide an equation that will provide only those 6 values, in addition to an even more general formula for finding the solutions to any arbitrary-sized matrix (b^n=1) because my professor likes to be nitpicky.
So, I'm not asking you to solve the entire physics problem for me, but I'm not a math major and I don't know this little part of complex analysis. So can someone help me get a general solution for b^n=1 with steps? Once I get a general solution for that, I can easily get the more specific solution for b^6=1.
*If you just want a "summary", just read the very last part. The rest is a little bit of background in case the origin of my problem might be of assistance.
3 Comments • Newest first
[quote=Obscene]17 year old basiler, reporting in with the answer.
b = cos(pi*n/3)+i*sin(pi*n/3), n = integers between 0 and 5 inclusive.
EDIT: Didn't realize he said show work. Anytime you have x^n=a, the answers all lie on a complex unit circle multiplied by a scalar a^(1/n). Also, when n is an integer, the answers are evently spaced (angularly), starting with theta=0. So from that point, divide 2pi by n. All multiples of this value in [0,2pi) are answers. Therefore the form is written x = a^(1/n) * [cos(2pi/n*k)+i*sin(2pi/n*k)].[/quote]
Awesome, thanks. Your answer is much more general and cleaner than the one that I managed to scrape up. I converted n to 2(N+1) and ended up with e^(i*pi*k/(N+1)), but I don't think that covers odd values of n. And it certainly doesn't cover the arbitrary value a. I think my professor will like your generalized form better.
I don't start complex analysis until next semester, so I was caught off-guard when I had to do this sort of thing for physics.
[quote=CrayonScribble]Honestly, how about you think before posting this on a children's forum with people mainly aged 13-17?
There are so many better options such as i.e asking a friend in your class, asking a TA, asking an older student or just reading the textbook a couple more times.
The only reasons I can fathom of you posting this on Basil is that you are mentally incapable of making sound judgements about the demography of a forum based for teenagers the majority of which have not gone to university and even less likely to study physics OR you are boasting; neither of these options are respectable in my opinion.[/quote]
17 year old basiler, reporting in with the answer.
b = cos(pi*n/3)+i*sin(pi*n/3), n = integers between 0 and 5 inclusive.
EDIT: Didn't realize he said show work. Anytime you have x^n=a, the answers all lie on a complex unit circle multiplied by a scalar a^(1/n). Also, when n is an integer, the answers are evently spaced (angularly), starting with theta=0. So from that point, divide 2pi by n. All multiples of this value in [0,2pi) are answers. Therefore the form is written x = a^(1/n) * [cos(2pi/n*k)+i*sin(2pi/n*k)].
Honestly, how about you think before posting this on a children's forum with people mainly aged 13-17?
There are so many better options such as i.e asking a friend in your class, asking a TA, asking an older student or just reading the textbook a couple more times.
The only reasons I can fathom of you posting this on Basil is that you are mentally incapable of making sound judgements about the demography of a forum based for teenagers the majority of which have not gone to university and even less likely to study physics OR you are boasting; neither of these options are respectable in my opinion.