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terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.5You 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?

ghazi
 one year ago
Best ResponseYou've already chosen the best response.18 (cos 216 + i sin 216) =\[8e^{i 216}\] rest you can do ;)

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

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

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.5Okay, 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...

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.5Let'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?

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.5In fact, equal to \[\large \cos (\theta + 2k\pi) + i \sin (\theta + 2k\pi)\] for any integer k. Catch me so far?

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.5But let's use degrees, it seems the situation calls for it... \[\cos \theta + i \sin \theta = \cos (\theta + 360k^o) + i \sin (\theta + 360k^o\]

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.5So 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...

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

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.5Could you do it from here? Could you post your answers, too?

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

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.5Applying the rule, we get... \[\huge 8^{\frac{1}{3}}(\cos \frac{216}{3}^o + i \sin \frac{216}{3}^o)\] \[\large =2(\cos 72^o + i \sin 72^o)\]

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.5However, we have to also consider the cube root of \[\large 8[\cos (216+360)^o + i \sin (216+360)^o]\] Since \[\large 8[\cos \ 216^o + i \sin \ 216^o]=\large 8[\cos (216+360)^o + i \sin (216+360)^o]\]

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.5applying the rule again, we get \[\huge 8^{\frac{1}{3}}(\cos \frac{216+360}{3}^o + i \sin \frac{216+360}{3}^o)\] \[\large = 2(\cos \ 192 + i \sin \ 192)\]

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.5However 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 \[\large 8[\cos \ 216^o + i \sin \ 216^o]=\large 8[\cos (216+720)^o + i \sin (216+720)^o]\]

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.5Applying the rule yet again, we get\[\huge 8^{\frac{1}{3}}(\cos \frac{216+720}{3}^o + i \sin \frac{216+720}{3}^o)\]\[\large = 2(\cos \ 312+ i \sin \ 312)\]

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.5Et 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:

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.5Consider\[\large 8[\cos (216+3(360))^o + i \sin (216+3(360))^o]\]\[=\large 8[\cos (216+1080)^o + i \sin (216+1080)^o]\] Applying the rule (yet again)...

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.5We get \[\huge 8^{\frac{1}{3}}(\cos \frac{216+1080}{3}^o + i \sin \frac{216+1080}{3}^o)\] \[\large = 2(\cos \ 432+ i \sin \ 432)\]

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.5But 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.

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.5Anyway, 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

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.5And that's it, I have to go now, have fun with Trig :) . Terence out
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