My Math Forum Complex Numbers Question

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 June 7th, 2014, 05:17 AM #1 Newbie   Joined: Apr 2012 Posts: 9 Thanks: 0 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.
 June 7th, 2014, 05:36 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,445 Thanks: 2499 Math Focus: Mainly analysis and algebra 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 $$z= 3e^{\frac23 n\pi i}$$ Thanks from lilstef
 June 16th, 2014, 06:27 AM #3 Newbie   Joined: Jun 2014 From: Parys Posts: 1 Thanks: 0 Is my try bad or not? »z^3=27i »z^3=3^3i therefore »z=3i.
 June 16th, 2014, 07:25 AM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,445 Thanks: 2499 Math Focus: Mainly analysis and algebra 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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