Additional Concepts
-$$x ^ { 3 } = - 27 i$$
A) $3 \left( \cos 90 ^ { \circ } + i \sin 90 ^ { \circ } \right) , 3 \left( \cos 210 ^ { \circ } + i \sin 210 ^ { \circ } \right) , 3 \left( \cos 330 ^ { \circ } + i \sin 330 ^ { \circ } \right)$
B) $3 \left( \cos 210 ^ { \circ } + i \sin 210 ^ { \circ } \right) , 3 \left( \cos 270 ^ { \circ } + i \sin 270 ^ { \circ } \right) , 3 \left( \cos 330 ^ { \circ } + i \sin 330 ^ { \circ } \right)$
C) $3 \left( \cos 30 ^ { \circ } + i \sin 30 ^ { \circ } \right) , 3 \left( \cos 60 ^ { \circ } + i \sin 60 ^ { \circ } \right) , 3 \left( \cos 90 ^ { \circ } + i \sin 90 ^ { \circ } \right)$
D) $3 \left( \cos 90 ^ { \circ } + i \sin 90 ^ { \circ } \right) , 3 \left( \cos 180 ^ { \circ } + i \sin 180 ^ { \circ } \right) , 3 \left( \cos 270 ^ { \circ } + i \sin 270 ^ { \circ } \right)$

Question

Quizplus · Accepted Answer

The equation $x^3 = -27i$ can be solved using De Moivre's theorem, which states that for any complex number $x = r(\cos \theta + i\sin \theta)$, its $n$th roots are given by $r^{1/n}(\cos(\frac{\theta + 2k\pi}{n}) + i\sin(\frac{\theta + 2k\pi}{n}))$, where $k = 0, 1, ..., n-1$. For $x^3 = -27i$, we have $r = 27$ and $\theta = 270^\circ$ (since $-i$ corresponds to the angle $270^\circ$ on the complex plane). The cube roots ($n = 3$) of this number are found by dividing the angle by 3 and adding $120^\circ$ (or $2\pi/3$ radians) for each subsequent root. This gives angles of $90^\circ$, $210^\circ$, and $330^\circ$, all with a magnitude of $3$ (since $27^{1/3} = 3$). Therefore, the correct roots are $3(\cos 90^\circ + i\sin 90^\circ)$, $3(\cos 210^\circ + i\sin 210^\circ)$, and $3(\cos 330^\circ + i\sin 330^\circ)$, matching choice A.

Additional Concepts - x3=−27ix ^ { 3 } = - 27 ix3=−27i

Additional Concepts
- $x ^ { 3 } = - 27 i$