We
already
know from Algebra II that every equation of power 2 has 2 solutions and
every
equation of power 3 has 3 solutions etc, etc ad nauseum. But up
until
now, we have only one solution for the equation x^{3} =
8.
The only one you could find was x =2. Where are the other
two?
We know they exist, but how do you find them. If we extend De
Moivre's
to find roots, suprise!! We can find the missing roots!!
How
you might ask? Good question! I guess I will show
you! Hang
on to your hat and off we go!

Important definition ( This means pay
attention!)

The n nth roots of z = r cis 0 are:

z^{1/n} = r^{1/n} cis ( 0/n
+ k ^{.} 360^{o} /n) for k = 0, 1, 2, 3, ... n-1

Say what? What are you trying to tell me? Here's
the deal! Say you want to find the 3 cubed roots of 8. That
is x^{3} = 8. Change 8 into r cis 0.
You remember that right!

8 is on the x-axis so r cis 0 = 8 cis 0^{o}.
Since we are working with the third power, k will equal 0, 1, and
2. Look at the formula! The highest k goes to is n-1.
Tada! Now plug and chug! One at a time.

2cos240 + 2i sin 240 = -1 - i\/ 3 . Another
imaginary number. Hey, I seem to recall something about these
guys showing up in pairs? Do you?

So how do you go
about proving what you really have are the roots of 8. How do you
check? Any brilliant ideas? That's right, you multiply. What? You
heard me, Multiply.

Proof: 2 x 2 x 2 =
8 (
boy that was tough!)

__
__ __

Proof: (-1
-i\/ 3)(-1 -i \/ 3)(-1 -i \/3 ) You guessed it FOIL!

__
__

(-2 + 2i \/ 3)( -1
- i
\/ 3) = 8 (Ripleys believe it or not!)

__
__
__

Proof: (-1 + i \/
3)(-1 + i \/ 3)( -1 + i \/ 3) =

__
__

(-2 - 2i \/ 3)( -1
- i
\/ 3) = 8 ( believe it if you must)

Basically, if you can do a couple of these monsters, you are
doing pretty well.

Thus ends chapter 11 and our glorious trek through the
cyberspace of trigonometry. But stay tuned, it's not quite over. Up
next is what you have waited for with baited breath! The sample test.