Finding complex numbers using DeMoivre's Theorem
Find the cube roots of 21/2cis(pi/4)
21/2cis(pi/4+ 2pi k), for every integer k
By DeMoivre:
21/21/3cis((pi/4+ 2pi k)/3)321/6cis(pi/12+ 2/3pi *k)
Taking k=0,1,2 gives the three cube roots:z1= 21/6cis(pi/12) (k=0)z2= 21/6cis(3pi/4) (k=1)z3= 21/6cis(17pi/12) (k=2)
LR
Related Maths IB answers
All answers ▸What does differentiation actually mean?
A team of four is chosen from six married couples. If a husband and wife cannot both be on the team, in how many ways can the team be formed?
Why is there a "+C" term in the solution of every indefinite integral?
a) Let u=(2,3,-1) and w=(3,-1,p). Given that u is perpendicular to w, find the value of p. b)Let v=(1,q,5). Given that modulus v = sqrt(42), find the possible values of q.