Find Roots of Complex Numbers in Polar Form
-The complex cube roots of 8 (rectangular form)
A) $2 , - 1 + \sqrt { 3 } i , - 1 - \sqrt { 3 } i$
B) $2,1 + \sqrt { 3 } i , 1 - \sqrt { 3 } i$
C) $2,1 + \sqrt { 3 } i , - 1 - \sqrt { 3 } i$
D) $2 , - 1 - \sqrt { 3 } i , 1 - \sqrt { 3 } i$

Question

Quizplus · Accepted Answer

The complex cube roots of 8 can be found by converting 8 into polar form, which is $8(\cos(0) + i\sin(0))$, and then applying De Moivre's theorem. The roots are $2(\cos(\frac{2k\pi}{3}) + i\sin(\frac{2k\pi}{3}))$ for $k=0, 1, 2$. For $k=0$, the root is $2$. For $k=1$, the root is $-1 + \sqrt{3}i$, and for $k=2$, the root is $-1 - \sqrt{3}i$.

Find Roots of Complex Numbers in Polar Form -The Complex 2,−1+3i,−1−3i2 , - 1 + \sqrt { 3 } i , - 1 - \sqrt { 3 } i2,−1+3​i,−1−3​i

Find Roots of Complex Numbers in Polar Form
-The Complex $2 , - 1 + \sqrt { 3 } i , - 1 - \sqrt { 3 } i$