Find all zeros of the polynomial P(x)= x^6 - 1. Enter roots as a list of complex numbers, such as -11,48,3-4i,5+i.

Answer 1

Noting that P(X)=0 is equivalent to x^6=1, so the question is asking for the six 6th roots of 1.

(x^n=a+bi will always have n solutions, given by the n nth roots of a+bi.)

To find all the real and imaginary roots of a complex number a+bi, one should begin by representing the number in its polar form r*(cos(theta) + i*sin(theta)) = r*cis(theta), where r=sqrt(a^2+b^2) and theta is given by arctan(b/a) if a>0, pi+arctan(b/a) if a<0, pi/2 if a=0, b>0, or 3pi/2 if a=0, b<0.

(If you do not know how to plot imaginary numbers in the complex plane, and determine r and theta, then we could discuss how to do so in a tutoring session.)

Having determined the polar form of the complex number a+bi, the corresponding nth roots of a+bi will be given by r^(1/n) * cis((theta+2pi*k) / n), for k=0,1,...,n-1.

(This formulas is presented here without justification but, again, such justification could be discussed in a session.)

In the current example, note that in the complex plane the point corresponding to a+bi = 1+0i corresponds to

(r, theta) = (1,0). So the six roots of 1+0i will be given by: 1^(1/6)*cis((0+2pi*k)/6) = cis(pi*k/3), for k=0,1,2,3,4,5.

cis(0*pi/3) = 1+0i = 1,

cis(1*pi/3) = 1/2 + sqrt(3)/2*i = i,

cis(2*pi/3) = -1/2+sqrt(3)/2*i,

cis(3*pi/3) = -1+0*i = -1,

cis(4*pi/3) = -1/2-sqrt(3)/2*i,

cis(5*pi/3) = 1/2-sqrt(3)/2*i.