Find Roots of Complex Numbers in Polar Form
-The complex square roots of $100 \left( \cos 210 ^ { \circ } + i \sin 210 ^ { \circ } \right)$ (polar form)
A) $10 \left( \cos 105 ^ { \circ } + i \sin 105 ^ { \circ } \right) , 10 \left( \cos 285 ^ { \circ } + i \sin 285 ^ { \circ } \right)$
B) $10 \left( \cos 210 ^ { \circ } + i \sin 210 ^ { \circ } \right) , 10 \left( \cos 195 ^ { \circ } + i \sin 195 ^ { \circ } \right)$
C) $10 \left( \cos 105 ^ { \circ } + i \sin 105 ^ { \circ } \right) , 195 \left( \cos 285 ^ { \circ } + i \sin 285 ^ { \circ } \right)$
D) $10 \left( \cos 210 ^ { \circ } + i \sin 210 ^ { \circ } \right) , - 10 \left( \cos 195 ^ { \circ } + i \sin 195 ^ { \circ } \right)$

Question

Quizplus · Accepted Answer

To find the square roots of a complex number in polar form, $r(\cos \theta + i\sin \theta)$, we use the formula for the nth root: $\sqrt[n]{r}(\cos \frac{\theta + 2k\pi}{n} + i\sin \frac{\theta + 2k\pi}{n})$, where $k = 0, 1, ..., n-1$. For square roots ($n=2$), and given $r = 100$ and $\theta = 210^\circ$, we find the roots by taking $\sqrt[2]{100} = 10$ and calculating the angles as $\frac{210^\circ + 2k\pi}{2}$ for $k=0$ and $k=1$, resulting in angles of $105^\circ$ and $285^\circ$.

Find Roots of Complex Numbers in Polar Form -The Complex 100(cos⁡210∘+isin⁡210∘)100 \left( \cos 210 ^ { \circ } + i \sin 210 ^ { \circ } \right)100(cos210∘+isin210∘)

Find Roots of Complex Numbers in Polar Form
-The Complex $100 \left( \cos 210 ^ { \circ } + i \sin 210 ^ { \circ } \right)$