## Online Course Discussion Forum

### MCIII Algebra Week 2 help

Here are some brief hints to get you started.

2.18: Note that z is a 7th root of unity (z^7 = 1). So this means that we can take any term in the given expression, and multiply (or divide) it by z^7, and the value will be unchanged, since we are essentially just multiplying by 1. See if you can apply this idea to some of the terms in the original summation and see if you get something that you recognize!

2.19: Remember that it's always easier to raise complex numbers to powers when we express them in the exponential form! Try to see if you can express z in exponential form (hint: try to solve the quadratic equation), then see what z^2000 and 1/z^2000 are and try to simplify.

2.21(c): Try to think about this more geometrically/conceptually. This is equivalent to |z-1| < |z+1|. What does this look like in the complex plane?

(Also, with this new form |z-1| < |z+1|, writing z as a+bi may still be helpful!)

2.22: Note that since w is one of the imaginary cube roots of unity, it is a root of x^3 - 1 = 0 that is not 1. In other words, w is a root of (x^3 - 1) / (x - 1) = x^2 + x + 1. Therefore, w^2 + w + 1 = 0. See how you can use this relationship to solve the problem.

2.23: Note that x is a cube root of unity that is not 1 (since it satisfies the same equation as in 2.22). So you can actually express the possible values of x in exponential form explicitly. Try to use this exponential form to get a sense for what x + 1/x, (x^2 + 1/x^2), (x^3 + 1/x^3), etc. are.

2.24: This problem is actually similar to an example problem which we weren't able to go over in class this past week (2.7). For this, you want to show that f(x) is divisible by the polynomial x^2 + x + 1, which means that you want to show that all roots of x^2 + x + 1 are roots of f(x). Again, x^2 + x + 1 has roots which are the complex cube roots of unity (seems like this equation has been showing up a lot in these problems!!). Try to proceed from here.

2.26: This is not on this week's homework; it will be on next week's homework. I will let you think about it for some more time!

Social networks