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 September 27th, 2016, 09:24 PM #1 Newbie   Joined: Apr 2016 From: Wonderland Posts: 10 Thanks: 0 Complex equation solving Find the roots of the equation (z^3 + 8 ) /(8 - z^3), giving each root in the form re^i(theta), where r > 0 and -180
 September 27th, 2016, 09:34 PM #2 Senior Member     Joined: Sep 2015 From: Southern California, USA Posts: 1,481 Thanks: 744 I assume you mean the equation $\dfrac{z^3 + 8}{8-z^3}=0$ The roots of this are the roots of $z^3+8=0$ now the trick is that we let $z=r e^{i\theta}$ $r^3 e^{i3 \theta} =-8$ $r=2$ $e^{i3\theta}=-1 = e^{i(\pi + 2k\pi)},~k\in \mathbb{Z}$ so $3\theta = \pi,~3\theta = 3\pi,~3\theta = -\pi$ $\theta = \dfrac{\pi}{3}, \pi, -\dfrac{\pi}{3}$ and finally $z = 2e^{i\frac{\pi}{3}}, 2e^{i\pi}, 2e^{-i \frac{\pi}{3}}$ Thanks from topsquark
September 28th, 2016, 01:13 AM   #3
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Quote:
 Originally Posted by romsek I assume you mean the equation $\dfrac{z^3 + 8}{8-z^3}=0$ The roots of this are the roots of $z^3+8=0$ now the trick is that we let $z=r e^{i\theta}$ $r^3 e^{i3 \theta} =-8$ $r=2$ $e^{i3\theta}=-1 = e^{i(\pi + 2k\pi)},~k\in \mathbb{Z}$ so $3\theta = \pi,~3\theta = 3\pi,~3\theta = -\pi$ $\theta = \dfrac{\pi}{3}, \pi, -\dfrac{\pi}{3}$ and finally $z = 2e^{i\frac{\pi}{3}}, 2e^{i\pi}, 2e^{-i \frac{\pi}{3}}$
Ok, I understood your method. What if the equation was $\dfrac{z^3 + 8}{8-z^3}=i$ instead? Will the roots still be the same?

September 28th, 2016, 07:06 AM   #4
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 Originally Posted by Appletree Ok, I understood your method. What if the equation was $\dfrac{z^3 + 8}{8-z^3}=i$ instead? Will the roots still be the same?
No, not at all. Why would they be?

you'd end up with $r^3 e^{i3\theta}=i+8$

and proceed from there. The algebra will be messier.

 November 4th, 2016, 01:05 PM #5 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,714 Thanks: 697 $\displaystyle \frac{z^3+ 8}{8- z^3}= i$ $\displaystyle z^3+ 8= 8i- iz^3$ $\displaystyle (1+ i)z^3= -8+ 8i$ $\displaystyle z^3= \frac{-8+ 8i}{1+ i}= \frac{-8+ 8i}{1+ i}\frac{1- i}{1- i}= 8i$ In "exponential form" or "polar form", writing $\displaystyle z= re^{i\theta}$, $\displaystyle r^3e^{3i\theta}= 8e^{i\pi/2}$ $\displaystyle r= 2$, $\displaystyle \theta= \frac{\pi}{6}$. $\displaystyle z=2e^{\frac{\pi}{6}}= \sqrt{3}+ i$ is one solution. Last edited by Country Boy; November 4th, 2016 at 01:09 PM.

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