## anonymous 3 years ago Find the cube roots of 8(cos 216° + i sin 216°).

1. anonymous

-7.978

2. terenzreignz

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?

3. anonymous

8 (cos 216 + i sin 216) =$8e^{i 216}$ rest you can do ;)

4. anonymous

Haha these are tough !! @terenzreignz

5. anonymous

$(8 e^{i 216} ) ^ {1/3}= 2 e^{i 72}$

6. terenzreignz

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...

7. terenzreignz

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?

8. anonymous

Okay I see

9. terenzreignz

In fact, equal to $\large \cos (\theta + 2k\pi) + i \sin (\theta + 2k\pi)$ for any integer k. Catch me so far?

10. terenzreignz

But let's use degrees, it seems the situation calls for it... $\cos \theta + i \sin \theta = \cos (\theta + 360k^o) + i \sin (\theta + 360k^o$

11. terenzreignz

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...

12. anonymous

Thank you @terenzreignz ! That makes a whole lot more sense now

13. terenzreignz

Could you do it from here? Could you post your answers, too?

14. anonymous

Yea Ill post the answer I got when I finish !

15. terenzreignz

Applying the rule, we get... $\huge 8^{\frac{1}{3}}(\cos \frac{216}{3}^o + i \sin \frac{216}{3}^o)$ $\large =2(\cos 72^o + i \sin 72^o)$

16. terenzreignz

However, we have to also consider the cube root of $\large 8[\cos (216+360)^o + i \sin (216+360)^o]$ Since $\large 8[\cos \ 216^o + i \sin \ 216^o]=\large 8[\cos (216+360)^o + i \sin (216+360)^o]$

17. terenzreignz

applying the rule again, we get $\huge 8^{\frac{1}{3}}(\cos \frac{216+360}{3}^o + i \sin \frac{216+360}{3}^o)$ $\large = 2(\cos \ 192 + i \sin \ 192)$

18. terenzreignz

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 $\large 8[\cos \ 216^o + i \sin \ 216^o]=\large 8[\cos (216+720)^o + i \sin (216+720)^o]$

19. terenzreignz

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)$

20. terenzreignz

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:

21. terenzreignz

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)...

22. terenzreignz

We get $\huge 8^{\frac{1}{3}}(\cos \frac{216+1080}{3}^o + i \sin \frac{216+1080}{3}^o)$ $\large = 2(\cos \ 432+ i \sin \ 432)$

23. terenzreignz

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.

24. terenzreignz

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

25. terenzreignz

And that's it, I have to go now, have fun with Trig :) --.-- Terence out