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

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

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

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

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

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

terenzreignz
 2 years 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
 2 years 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
 2 years 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
 2 years 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
 2 years 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
 2 years 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
 2 years 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
 2 years 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
 2 years 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
 2 years 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
 2 years 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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