A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 years ago
Find the cube roots of 8(cos 216° + i sin 216°).
anonymous
 3 years ago
Find the cube roots of 8(cos 216° + i sin 216°).

This Question is Closed

terenzreignz
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.18 (cos 216 + i sin 216) =\[8e^{i 216}\] rest you can do ;)

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

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

terenzreignz
 3 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
 3 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
 3 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
 3 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
 3 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...

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

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

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

terenzreignz
 3 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
 3 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
 3 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
 3 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
 3 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
 3 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
 3 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
 3 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
 3 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
 3 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
 3 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
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.