lee wrote:(i) given that w=1+2i, express w2, w3, and w4 in the form a+bi.
(ii) given that w is a root of the equation z4+pz3+qz2-6z+65=0, find the values of p and q.
(iii) write down a second root of the equation.
(iv) find the other 2 roots of the equation.
(i) do the powers: w^2 = (1+2i)^2 = (1+2i)(1+2i) = 1 + 4i + 4i^2 = 1 + 4i - 4 = -3 + 4i
do the same for the others
(ii) since 1+2i is a root then 1-2i is a root too
so do the synthetic div w/ each root: http://www.purplemath.com/modules/synthdiv.htm
put the remainder =0
i get (68-3q+5p)+(-4+4q-2p)i=0=0*1+0*i
this gives a linear system: 5p-3q=-68, -2p+4q=4
(iii) since 1+2i is a root so is 1-2i
divide these out
this leaves a quadratic
if U get stuck plz write back w/ your working