We start by considering the given equation:
(z−100)3+1000=0This can be rewritten as:
(z−100)3=−1000Noting that -1000 is the same as
−103, we get:
(z−100)3=(−10)3We can now take the cube root of both sides:
z−100=−10z−100=−10ωz−100=−10ω2Here,
ω and
ω2 are the other two primitive cube roots of unity. Solving these equations for
z, we get:
For
z−100=−10 :
z=90For
z−100=−10ω :
z=100−10ωFor
z−100=−10ω2 :
z=100−10ω2 To find the expressions in the given options, recall the properties of the cube roots of unity. If
ω is a primitive cube root of unity, then we have:
ω3=11+ω+ω2=0This means we can express 1 as:
1=−ω−ω2Substituting
1=−ω−ω2 in the solutions we found:
For
z=100−10ω :
z=10(10−ω)For
z=100−10ω2.
z=10(10−ω2)So, the solutions are:
10(10−ω),10(10−ω2), and 90