Solve: z² + z + 1 = 0:
Az = ±i
Bz = ±1
Cz = (-1 ± i√3)/2 (complex cube roots of unity other than 1)
DNo solution
Answer & Solution
Correct answer: C. z = (-1 ± i√3)/2 (complex cube roots of unity other than 1)
z = (-1 ± √(1 - 4))/2 = (-1 ± √(-3))/2 = (-1 ± i√3)/2. These are ω and ω² (complex cube roots of unity).
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?