Question

Find the cube roots of 8(cos 216° + i sin 216°).

Answer 1

-7.978

Answer 2

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?

Answer 3

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

Answer 4

Haha these are tough !! @terenzreignz

Answer 5

\[(8 e^{i 216} ) ^ {1/3}= 2 e^{i 72}\]

Answer 6

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...

Answer 7

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?

Answer 8

Okay I see

Answer 9

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

Answer 10

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

Answer 11

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...

Answer 12

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

Answer 13

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

Answer 14

Yea Ill post the answer I got when I finish !

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

Applying the rule yet again, we get\[\huge 8^{\frac{1}{3}}(\cos \frac{216+720}{3}^o + i \sin \frac{216+720}{3}^o)\]\[\large = 2(\cos \ 312+ i \sin \ 312)\]

Answer 20

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:

Answer 21

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)...

Answer 22

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

Answer 23

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.

Answer 24

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

Answer 25

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