Home › JEE Advanced › Mathematics › Complex Numbers › If $\omega$ is a non-real cube root of unity, fi…
If $\omega$ is a non-real cube root of unity, find the value of $1 + \omega + \omega^2$.
A$0$
B$3$
C$-1$
D$1$
Answer & Solution
Correct answer: A. $0$
The three cube roots of unity are $1, \omega, \omega^2$, and they are roots of $z^3 - 1 = (z-1)(z^2 + z + 1) = 0$. Vieta's formulas on $z^2 + z + 1$ give $\omega + \omega^2 = -1$, so $1 + \omega + \omega^2 = 0$. Equivalently, the sum of *all* $n$-th roots of unity is zero for $n \ge 2$.
Related questions
Using De Moivre's theorem, the complex roots of $z^3=1$ are:If $\alpha$ and $\beta$ are roots of $z^2+4z+8=0$, then $\dfrac{\alpha+\beta+4i}{\alpha\beThe locus $\left|\dfrac{z-2}{z+3i}\right|=4$ simplifies to which equation?Simplify i raised to the power 27. What is the result?Multiply (2 + 3i) by (4 - i). What is the product in standard form?How is the complex conjugate of a complex number obtained?Add the complex numbers 5 - 2i and 3 + 7i. What is the sum?In the complex number 7 + 4i, which part is the real part?