1. Recall that the regular polygon in C is a polygon such that all angles at its vertices are equal and all sides have the same length. Let 21, Z2, ..., En be the vertices of the regular polygon in C on the unit circle, i.e., |z,| = 1, n = 3, 4,5, ... . Prove that 21 + 22 + ...+ 2n = 0. Hint. You can use properties of e and the geometric progression formula for complex numbers: if z # 1 and n E N, n 2 2, then 1+z+2"+...+ zn-1_ 1-2" 1 - z 2. (Quaternions). Let 1 = (1, 0, 0, 0) = 1, i = (0, 1, 0, 0), j = (0, 0, 1, 0) and k = (0, 0, 0, 1). Let H = R' = {q = (To, X1, 12, 13) = (To1 + x1i + 12j + 3k) : To, X1, 12, 13 ( R). Addition of vectors in H is the same as addition of vectors in R4. The table of multiplication is given below: * 1 i k 1 1 Li k i i -1 k -j j j -k -1 i k k j -i where the left column represents the left factor of the product and the upper row represents the right factor: i? = j? = k2 = -1; i*j = k, j*k = i, k*i = j; i*k = -j, etc. For example, (1 + 2j) * (i + k) = 1*i+1*1* k +2(j *i) + 2(j * k) = it k+2(-k) + 2i = 3i - k. Given q = Tol + Cli + Taj + 13k # 0, define |q) , q and find q , that is, find such quaternion q that q q = qq ] = 1. Notice that the quaternion multiplication is not commutative. Hence, you have to verify both identities q q = qq l = 1 for your choice of q-]. You can assume that the axioms Al-A4, M2-M4 and D hold for the quaternions relative to the standard operation of addition "+" and the multiplication "*" which we introduced above