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September 27th, 2016, 09:24 PM   #1
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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?
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September 27th, 2016, 09:34 PM   #2
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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

$z = 2e^{i\frac{\pi}{3}}, 2e^{i\pi}, 2e^{-i \frac{\pi}{3}}$
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September 28th, 2016, 01:13 AM   #3
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Quote:
Originally Posted by romsek View Post
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?
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September 28th, 2016, 07:06 AM   #4
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Originally Posted by Appletree View Post
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.
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November 4th, 2016, 01:05 PM   #5
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$\displaystyle \frac{z^3+ 8}{8- z^3}= i$
$\displaystyle z^3+ 8= 8i- iz^3$
$\displaystyle (1+ i)z^3= -8+ 8i$
$\displaystyle z^3= \frac{-8+ 8i}{1+ i}= \frac{-8+ 8i}{1+ i}\frac{1- i}{1- i}= 8i$

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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