Question

If $\omega \ne 1$ is a cube root of unity, then roots of $(x-2i{)}^3+i=0$ are

• A. $i,i,i$
• B. $3i,2i+i\omega ,2i+i{\omega }^2$
• C. $2i,2i,2i$
• D. $i,i+\omega ,i+{\omega }^2$

Answer 1

Answer: (B)

$3i,2i+i\omega ,2i+i{\omega }^2$

${(x-2i)}^3+i=0$
$\Rightarrow {\left(\frac{x-2i}{i}\right)}^3=1$
On substituting $z=\frac{x-2i}{i}$, we get
$z^3=1$
Therefore, $z$ is the cube root of unity.
$\Rightarrow z=1,w,w^2$
$\Rightarrow \frac{x-2i}{i}=1,w,w^2$
$\Rightarrow x=3i,2i+iw,2i+i{w}^2$
Ans: B