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terenzreignzBest ResponseYou've already chosen the best response.3
You again? LOL, let's get down to business :D First off, I'm going to write \[\large r(\cos \theta + i \sin \theta) = r \ cis \theta\] It's shorter that way :P Ready?
 one year ago

ghaziBest ResponseYou've already chosen the best response.1
8 (cos 216 + i sin 216) =\[8e^{i 216}\] rest you can do ;)
 one year ago

jazzieBest ResponseYou've already chosen the best response.0
Haha these are tough !! @terenzreignz
 one year ago

ghaziBest ResponseYou've already chosen the best response.1
\[(8 e^{i 216} ) ^ {1/3}= 2 e^{i 72}\]
 one year ago

terenzreignzBest ResponseYou've already chosen the best response.3
Okay, since you're awesome (^.^), I'll not use shortcuts... So, if we have \[\large z = r (\cos \theta + i \sin \theta)\] In general, we get \[\large z^p=r^p(\cos \ p\theta + i \sin \ p\theta)\] I say in general, because in the case of fractional exponents (like taking the cube root, for instance), there's a minor catch...
 one year ago

terenzreignzBest ResponseYou've already chosen the best response.3
Let's look at this part first, for instance: \[\cos \theta + i \sin \theta\] This is equal to \[\large \cos (\theta + 2 \pi) + i \sin (\theta + 2\pi)\] right?
 one year ago

terenzreignzBest ResponseYou've already chosen the best response.3
In fact, equal to \[\large \cos (\theta + 2k\pi) + i \sin (\theta + 2k\pi)\] for any integer k. Catch me so far?
 one year ago

terenzreignzBest ResponseYou've already chosen the best response.3
But let's use degrees, it seems the situation calls for it... \[\cos \theta + i \sin \theta = \cos (\theta + 360k^o) + i \sin (\theta + 360k^o\]
 one year ago

terenzreignzBest ResponseYou've already chosen the best response.3
So back to the question: Cube roots of \[\large 8(\cos \ 216^o + i \sin \ 216^o)\] Now use this rule \[\large z^p=r^p(\cos \ p\theta + i \sin \ p\theta)\] and since we're taking the cube root, take p = 1/3... go ahead now...
 one year ago

jazzieBest ResponseYou've already chosen the best response.0
Thank you @terenzreignz ! That makes a whole lot more sense now
 one year ago

terenzreignzBest ResponseYou've already chosen the best response.3
Could you do it from here? Could you post your answers, too?
 one year ago

jazzieBest ResponseYou've already chosen the best response.0
Yea Ill post the answer I got when I finish !
 one year ago

terenzreignzBest ResponseYou've already chosen the best response.3
Applying the rule, we get... \[\huge 8^{\frac{1}{3}}(\cos \frac{216}{3}^o + i \sin \frac{216}{3}^o)\] \[\large =2(\cos 72^o + i \sin 72^o)\]
 one year ago

terenzreignzBest ResponseYou've already chosen the best response.3
However, we have to also consider the cube root of \[\large 8[\cos (216+360)^o + i \sin (216+360)^o]\] Since \[\large 8[\cos \ 216^o + i \sin \ 216^o]=\large 8[\cos (216+360)^o + i \sin (216+360)^o]\]
 one year ago

terenzreignzBest ResponseYou've already chosen the best response.3
applying the rule again, we get \[\huge 8^{\frac{1}{3}}(\cos \frac{216+360}{3}^o + i \sin \frac{216+360}{3}^o)\] \[\large = 2(\cos \ 192 + i \sin \ 192)\]
 one year ago

terenzreignzBest ResponseYou've already chosen the best response.3
However still, we have to also consider the cube root of \[\large 8[\cos (216+2(360))^o + i \sin (216+2(360))^o]\]\[=\large 8[\cos (216+720)^o + i \sin (216+720)^o]\] Since \[\large 8[\cos \ 216^o + i \sin \ 216^o]=\large 8[\cos (216+720)^o + i \sin (216+720)^o]\]
 one year ago

terenzreignzBest ResponseYou've already chosen the best response.3
Applying the rule yet again, we get\[\huge 8^{\frac{1}{3}}(\cos \frac{216+720}{3}^o + i \sin \frac{216+720}{3}^o)\]\[\large = 2(\cos \ 312+ i \sin \ 312)\]
 one year ago

terenzreignzBest ResponseYou've already chosen the best response.3
Et voila! Your three cube roots. I'll show you a shortcut for this, but first, let's see what happens if continue this process... namely that we take the cube root, only add 360 degrees more to the angle:
 one year ago

terenzreignzBest ResponseYou've already chosen the best response.3
Consider\[\large 8[\cos (216+3(360))^o + i \sin (216+3(360))^o]\]\[=\large 8[\cos (216+1080)^o + i \sin (216+1080)^o]\] Applying the rule (yet again)...
 one year ago

terenzreignzBest ResponseYou've already chosen the best response.3
We get \[\huge 8^{\frac{1}{3}}(\cos \frac{216+1080}{3}^o + i \sin \frac{216+1080}{3}^o)\] \[\large = 2(\cos \ 432+ i \sin \ 432)\]
 one year ago

terenzreignzBest ResponseYou've already chosen the best response.3
But 432 is already past 360, so it's the same if we subtract 360 from it, and we'd get \[\large =2(\cos 72^o + i \sin 72^o)\] Which was the first answer we got, so we really only have 3 cube roots.
 one year ago

terenzreignzBest ResponseYou've already chosen the best response.3
Anyway, a shortcut for this, when getting the nth roots of a complex number in polar form, you just apply the rule directly, and only once... \[\huge 8^{\frac{1}{3}}(\cos \frac{216}{3}^o + i \sin \frac{216}{3}^o)\] \[\large =2(\cos 72^o + i \sin 72^o)\] And just keep adding \[\frac{360}{n}\] degrees to the angle until you have n distinct answers. In this case, n = 3, so keep addint 360/3 = 120 to the angle until you have three cube roots... The angles are 72 72 + 120 = 192 72 + 120 + 120 = 312
 one year ago

terenzreignzBest ResponseYou've already chosen the best response.3
And that's it, I have to go now, have fun with Trig :) . Terence out
 one year ago
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