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

Question

shubhamsrg · Answer

Euler theorem to the aid! B|

hartnn · Answer

8(cos 216° + i sin 216°).

anonymous · Answer

Is polar form a + bi (standard form)?

hartnn · Answer

can you write 8(cos 216° + i sin 216°) in polar form ?
r (angle) theta

terenzreignz · Answer

Is it not already in polar form?

hartnn · Answer

ok, i meant 8 <216

terenzreignz · Answer

Although, for the sake of ease of typing, might we set
r(cos A + isin A) = r cis A

hartnn · Answer

so, in such a case, to find cube root, first just take the cube root of magnitude
so, sube root of 8 =... ?

hartnn · Answer

*cube

anonymous · Answer

is 2

hartnn · Answer

and just do 1/3 of the angle..

terenzreignz · Answer

Use this formula....

$$\large \left[ r(\cos \theta + i \sin \theta) \right]^p=r^p(\cos \ p \theta + i \sin \ p \theta)$$

terenzreignz · Answer

And remember that taking the cube root is just raising to a fractional exponent, 1/3.

anonymous · Answer

Oh

hartnn · Answer

i was typing that only.. p=1/3

anonymous · Answer

So at k=0, 2(cos pi/6 + isin pi/6)

terenzreignz · Answer

But... with fractional exponents, it's not that simple... but it isn't too complicated neither :)
There's a slight twist XD

anonymous · Answer

And at k=1, 2(cos 5pi/6 + isin 5pi/6)

terenzreignz · Answer

Wait a minute, how did you get pi/6?

anonymous · Answer

@hartnn  Please give @terenzreignz a medal for me! Thanks to both of you for your help, I got it. :)

hartnn · Answer

i gave him/her way before you thought...

terenzreignz · Answer

I think that's been done :)
But where did you get pi/6 ?

@hartnn 
I'm Terence --> I'm a boy XD

hartnn · Answer

noted.