Find all cube roots of 8i.
Express each root in the form a+bi, where "a" and "b" are real numbers.
+118690 There are 3 roots each 2pi/3 = 4pi/6 apart
\(\sqrt[3]{8i}=-2i\)
-2i has a anticlockwise angle of \(\frac{3\pi}{2}=\frac{9\pi}{6}\\ \)
\(\frac{9\pi}{6}-\frac{4\pi}{6}=\frac{5\pi}{6}\\ \frac{5\pi}{6}-\frac{4\pi}{6}=\frac{9\pi}{6}-\frac{4\pi}{6}=\frac{5\pi}{6}\\~\\ \text{so the 3 cubic roots are }\quad \\ 2e^{(\frac{\pi}{6})i},\quad 2e^{(\frac{5\pi}{6})i},\quad 2e^{(\frac{3\pi}{2})i}\\ \sqrt[3]{8i}=2(cos\frac{\pi}{6}+isin\frac{\pi}{6}),\;2(cos\frac{5\pi}{6}+isin\frac{5\pi}{6}),\;2(cos\frac{3\pi}{2}+isin\frac{3\pi}{2})\\ etc\)
checked:
https://www.wolframalpha.com/input/?i2d=true&i=Surd%5B8i%2C3%5D
LaTex:
\frac{9\pi}{6}-\frac{4\pi}{6}=\frac{5\pi}{6}\\
\frac{5\pi}{6}-\frac{4\pi}{6}=\frac{9\pi}{6}-\frac{4\pi}{6}=\frac{5\pi}{6}\\~\\
\text{so the 3 cubic roots are }\quad \\
2e^{(\frac{\pi}{6})i},\quad 2e^{(\frac{5\pi}{6})i},\quad 2e^{(\frac{3\pi}{2})i}\\
\sqrt[3]{8i}=2(cos\frac{\pi}{6}+isin\frac{\pi}{6}),\;2(cos\frac{5\pi}{6}+isin\frac{5\pi}{6}),\;2(cos\frac{3\pi}{2}+isin\frac{3\pi}{2})\\
etc
