The value of (-i)^{1/3} is | vedclass

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(D) We need to find the cube roots of $-i$. In polar form,$-i = \cos(\frac{3\pi}{2}) + i\sin(\frac{3\pi}{2})$. The cube roots are given by $z_k = \cos

Vedclass · Accepted Answer

Both $B$ and $C$

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(D) We need to find the cube roots of $-i$. In polar form,$-i = \cos(\frac{3\pi}{2}) + i\sin(\frac{3\pi}{2})$. The cube roots are given by $z_k = \cos(\frac{3\pi/2 + 2k\pi}{3}) + i\sin(\frac{3\pi/2 + 2k\pi}{3})$ for $k = 0, 1, 2$. For $k=0$: $z_0 = \cos(\frac{\pi}{2}) + i\sin(\frac{\pi}{2}) = i$. For $k=1$: $z_1 = \cos(\frac{7\pi}{6}) + i\sin(\frac{7\pi}{6}) = -\frac{\sqrt{3}}{2} - \frac{1}{2}i = \frac{-\sqrt{3} - i}{2}$. For $k=2$: $z_2 = \cos(\frac{11\pi}{6}) + i\sin(\frac{11\pi}{6}) = \frac{\sqrt{3}}{2} - \frac{1}{2}i = \frac{\sqrt{3} - i}{2}$. Since both $\frac{-\sqrt{3} - i}{2}$ and $\frac{\sqrt{3} - i}{2}$ are roots of $(-i)^{1/3}$,the correct option is $D$.

List-$I$	List-$II$
$P.$ For each $z_k$ there exists a $z_j$ such that $z_k \cdot z_j = 1$	$1.$ True
$Q.$ There exists a $k \in \{1, 2, \ldots, 9\}$ such that $z_1 \cdot z = z_k$ has no solution $z$ in the set of complex numbers.	$2.$ False
$R.$ $\frac{\|1-z_1\|\|1-z_2\| \ldots \|1-z_9\|}{10}$ equals	$3.$ $1$
$S.$ $1 - \sum_{k=1}^9 \cos \left(\frac{2k\pi}{10}\right)$ equals	$4.$ $2$

The value of $(-i)^{1/3}$ is

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