Question

Find the complex cube roots of -8i. Help please!

Answer 1

hint:
we can write this:
\[\Large \begin{gathered}
- 8i = 8 \times \left( { - i} \right) = \hfill \\
= 8\left\{ {\cos \left( {\frac{{3\pi }}{2} + 2k\pi } \right) + i\sin \left( {\frac{{3\pi }}{2} + 2k\pi } \right)} \right\},\quad k \in \mathbb{Z} \hfill \\
\end{gathered} \]

Answer 2

Would you just replace K with 1,2,3,etc??

Answer 3

we have to take the cube roots of the right side, as below:
\[\large \sqrt[3]{8}\left\{ {\cos \left( {\frac{{3\pi }}{{2 \times 3}} + \frac{{2k\pi }}{3}} \right) + i\sin \left( {\frac{{3\pi }}{{2 \times 3}} + \frac{{2k\pi }}{3}} \right)} \right\},\quad k = 0,1,2\]

Answer 4

and we have to replace k with 0, 1, 2

Answer 5

for example, the first cube root, is given with k=0, namely:

Answer 6

\[\Large \begin{gathered}
2\left\{ {\cos \left( {\frac{{3\pi }}{{2 \times 3}}} \right) + i\sin \left( {\frac{{3\pi }}{{2 \times 3}}} \right)} \right\} = \hfill \\
= 2\left\{ {\cos \left( {\frac{\pi }{2}} \right) + i\sin \left( {\frac{\pi }{2}} \right)} \right\} = \hfill \\
= 2\left\{ {0 + i} \right\} = \hfill \\
= 2i \hfill \\
\end{gathered} \]