Find Roots of Complex Numbers in Polar Form
-The complex cube roots of $8 \mathrm { i }$ (rectangular form)
A) $- 2 \mathrm { i } , \sqrt { 3 } + \mathrm { i } , - \sqrt { 3 } + \mathrm { i }$
B) $- 2 \mathrm { i } , - \sqrt { 3 } - \mathrm { i } , \sqrt { 3 } - \mathrm { i }$
C) $2 \mathrm { i } , \sqrt { 3 } - \mathrm { i } , - \sqrt { 3 } - \mathrm { i }$
D) $2 \mathrm { i } , \sqrt { 3 } + \mathrm { i } , - \sqrt { 3 } + \mathrm { i }$

Question

Quizplus · Accepted Answer

The complex cube roots of $8 \mathrm { i }$ are given by $$8^{1/3} \left ( \cos \frac{ \pi}{6} + i \sin \frac{ \pi}{6} \right )$$$$2 \left ( \cos \frac{ \pi}{6} + i \sin \frac{ \pi}{6} \right )$$$$2 \left ( \frac{\sqrt{3}}{2} + \frac{1}{2} i \right )$$$$= \sqrt{3} + i$$The other two cube roots are obtained by multiplying this root by the cube roots of unity, which are $$w = \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3} \quad \text{and} \quad w^2 = \cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3}$$Therefore, the three cube roots of $8 \mathrm { i }$ are $$2 \left ( \cos \frac{ \pi}{6} + i \sin \frac{ \pi}{6} \right )$$$$2 \left ( \cos \frac{ \pi}{6} + i \sin \frac{ \pi}{6} \right ) w$$$$2 \left ( \cos \frac{ \pi}{6} + i \sin \frac{ \pi}{6} \right ) w^2$$$$= \sqrt{3} + i, \quad -\sqrt{3} + i, \quad -2i$$

Find Roots of Complex Numbers in Polar Form -The Complex 8i8 \mathrm { i }8i

Find Roots of Complex Numbers in Polar Form
-The Complex $8 \mathrm { i }$