The cube roots of 1 + i. Referring to the labels from the graph you selected above, give the exact roots. w0= w1= w2=

Answer 1

cube root of 1+i = a+bi

1+i = (a+bi)^3 = a^3 +3(a^2)(bi) + 3a(-b^2) + b^3(-i)

1+i = a^3-3ab^2 + 3a^2bi -ib^3

1 = a^3 -3ab^2

1 = 3a^2b -b^3

two equations two unknowns has 3 solutions for a and b, as they are 3 degree equations

a^3 -3ab^2 = 3a^2b - b^3

a(a^2 -3b^2)= b(3a^2 -b^2)

a+bi is the cube root of 1+i

or use De Moivre's Theorem that if r =r(cis) then r^n = rn(cis)

cis = cos(theta) + isin(theta)

3 cube roots of 1+i

=

(6th root of 2) times cis(.75pi)

(6th root of 2) times cis(17pi/12)

and

(6th root of 2) times cis(pi/12)

a+bi is the cube root where a and b have 3 different values

a= (6th root of 2)cos.75, b = (6th root of 2)sin.75

and

a = (6th root of 2)cos(17/pi/12), b = (6th root of 2)sin(17pi/12)

and

a = (6th root of 2)cos(pi/12), b = (6th root of 2)sin(pi/12)

Answer 2

Putting 1+i into cis form: r = sqrt(1+1) = sqrt(2) tan(θ) = 1, so θ = π/4

The principle cube root of rcis(θ) = r^1/3cis(θ/3) = 2^1/6cis(π/12)

The other two roots can be obtained by adding 2π/3 to the angle: Angles are 3π/4 and 17π/12 (or -7π/12)

They can be written in regular form by expanding cis(θ) to r(cosθ + isin(θ))