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June 7th, 2014, 05:17 AM   #1
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Complex Numbers Question

Hey guys, I never understood and learned the way to solve these kind of questions so I will ask for a bit of help before my final exam. Can you please solve this step by step for me? Thank you in advance!
Solve the equation z^3=27i and display the solutions in the complex number plane.
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June 7th, 2014, 05:36 AM   #2
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Let $z = re^{i\theta}$ then
$$z^3 = r^3e^{3i\theta} = 27e^{2n\pi i} \qquad \Longrightarrow \qquad \begin{cases}
r = 3 \\
3\theta = 2n\pi \\
\end{cases}$$
Thus
$$z= 3e^{\frac23 n\pi i}$$
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June 16th, 2014, 06:27 AM   #3
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Is my try bad or not?
»z^3=27i
»z^3=3^3i
therefore
»z=3i.
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June 16th, 2014, 07:25 AM   #4
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You haven't taken the cube-root of $i$. I didn't see the $i$ there in the first place!

$$z^3 = r^3e^{3i\theta} = 27e^{\left(2n+\frac{1}{2}\right)\pi i} \qquad \Longrightarrow \qquad \begin{cases}
r = 3 \\
3\theta = \left(4n+1\right)\frac{\pi}{2} \implies \theta = \left(4n+1\right)\frac{\pi}{6}\\
\end{cases}$$
Which gives
$$z = 3e^{i \frac{\pi}{6} },3e^{i \frac{5\pi}{6} },3e^{i \frac{9\pi}{6} }$$

Last edited by v8archie; June 16th, 2014 at 07:35 AM.
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