Question

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

Answer 1

I would just like the three formulas.

Answer 2

Unsimplified.

Answer 3

I usually try to derive them myself:
\[\Large z^3=8\text{cis}(216) =(r\text{cis}(\phi))^3\]

Now apply De Moivre:

\[\Large r^3\text{cis}(3\phi)=8\text{cis}(216°) \]

So that two complex numbers are identical, their argument and their magnitude has to be the same (or a multiply of 360° of each other:

\[\Large r^3=8 \]
and
\[\Large 3\phi=216°+n360°, \ n \in \mathbb{Z} \]

Answer 4

Umm...

Answer 5

I'm confused.

Answer 6

What is cis?

Answer 7

and phi?

Answer 8

An abbreviation for cos(alpha)+isin(alpha) made for lazy people like me (-: but it's official !

Answer 9

Oh, okay.

Answer 10

phi is an angle, you have to solve for the angle and then that gives you three solutions. you can solve for phi and then set n=0, n=1, n=2 that gives you the three solutions.

Answer 11

Now I'm very confused.

Answer 12

ok I am sorry, that was not my intend, to understand what I have done above you need to have a basic insight into Laws of De Moivre (complex number theorem)

And coefficient matching

Maybe the solution formula would be easier for you in that case, if the above gives you trouble.

But the concept is still very simple, all I do is comparing the LHS to the RHS and set things equal, such that the equal sign in the middle remains valid.

Answer 13

LHS?

Answer 14

What is that?

Answer 15

LHS= Left Hand Side
RHS= Right Hand Side

Answer 16

the left and the right of an equation.

Answer 17

Oh.

Answer 18

So what is the formula?

Answer 19

http://www.suitcaseofdreams.net/Roots_complex.htm

there it has a full introduction and also the solution formula for the n-th root of complex numbers.

Answer 20

This is so confusing!!!

Answer 21

Sorry.

Answer 22

maybe someone manages to give you another solution to this problem which you find less confusing, don't hesitate to reopen that question again if it remains unanswered.

Answer 23

Okay.