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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<theta<(include)180 Sorry about the poor formatting, but I got stuck halfway. I multiplied the denominator over, and then made the equation = 0. I got stuck since I had z = 2 or 2 which doesn't make sense. Any tips on solving? 
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 $\dfrac{z^3 + 8}{8z^3}=0$ 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 $z = 2e^{i\frac{\pi}{3}}, 2e^{i\pi}, 2e^{i \frac{\pi}{3}}$ 
September 28th, 2016, 01:13 AM  #3  
Newbie Joined: Apr 2016 From: Wonderland Posts: 10 Thanks: 0  Quote:
 
September 28th, 2016, 07:06 AM  #4 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,481 Thanks: 744  
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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