If w is a complex cube root of unity then (z+1)(z+w)(z-w-1)=
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Answer: among them 1 is real number and the other two are conjugate complex numbers and they are also known as imaginary cube roots of unity.
Step-by-step explanation: Then, cubing both sides we get, z3 = 1
or, z3 - 1 = 0
or, (z - 1)(z2 + z + 1) = 0
Therefore, either z - 1 = 0 i.e., z = 1 or, z2 + z + 1 = 0
Therefore, z = −1±12−4⋅1⋅1√2⋅1 = −1±−3√2 = -12 ± i√32
Therefore, the three cube roots of unity are
1, -12 + i√32 and -12 - i√32
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