Use De Moivre's theorem to find all of the solutions to the equation x^3-1=0.

Question

In the solution to this question:

1 = z = r^3(cos(3θ)+isin(3θ))

Then this implies that r=1 (or r= -1, but we can incorporate the latter case into our choice of angle). We then reduce the equation.

I do not understand how this implies that r=1. Could someone explain how this implies that? There is an entire trigonometric term multiplying the r^3, so wouldn't that term have some influence on the expression? What am I missing?

Answer 1

solving z³=1,,,,,,,,,,not z³=-1

all z can be represented z=re^iθ=r(cosθ+isinθ)=rcosθ+irsinθ=x+iy

the distance from the origin to point z is r = [(rcosθ)²+(rsinθ)²]^1/2

if r=1 distance is [cos²θ+sin²θ]^1/2=1,,,,,,,,z=cosθ+isinθ has distance from origin =1

update

really should proceed with z³=1 is same as z_k=(cos(0+2πk)+isin(0+2πk))^1/3 ,,,k=0,1,2

apply de moive theorem

note e^i2π/3=cos(2π/3)+isin(2π/3)=z₁ for example

and for z³=-1 is same as z_k=(cos(π+2πk)+isin(π+2πk))^1/3,,,apply de moive theorem

this works too.

substitute z=re^iθ,,,,r is positive real number

solving r³e^i3θ=1=1*(1+0i),,,,,,have r³=1 in real number system,,,r=1

also cos3θ=1 & sin3θ=0

3θ=2πk ,,,,k=0,1,2

roots of polynomial z³=1 are

z₁=e⁰

z₂=e^i2π/3

z3=e^i4π/3

check (z-1)(z-e^i2π/3)(z-e^i4π/3)=f(z)

z³-1=f(z)

note: for z³=-1

equation is r³e^i3θ=1=1*(-1+0i),,,,,r=1 still

but now cos3θ=(-)1 & sin3θ=0

root z₂=e^iπ=-1