Originally Posted by

**Mentia** Your #2 looks good!

I don't think your #3 is correct though. Lets take a look:

ArcTan(2/2) = Pi/4 = Theta

r = (2^2+2^2)^(1/2) = 8^(1/2)

n = 3

then z = 8^(1/6) * ( cos[ (Pi/4 + 2*Pi*k) / n ] + i*sin[ (Pi/4 + 2*Pi*k) / n ])

plug in 0, 1 and 2 for k and you get the answers (remember, you plug in values for k such that k = 0, 1, ..., n-1:

I am going to give you the decimal solutions, you can figure out the exact if you wish.

k = 0 -> z = 1.366025 + i*.366025

k = 1 -> z = -1 + i

k = 2 -> -.366025 - i*1.366025

so the answer to your question #3 is the root where k = 1:

z = -1 + i

raise that to the 3rd power and you'll find z^3 = 2 + 2i as expected.

As far as #1 is concerned, you do the same thing as I just did for #3, you find all the roots by plugging in all the k values. Note that k = 0, 1, ..., n - 1.

So if you want the 2nd root you use k = 0 or 1

if you want the 3rd root you use k = 0, 1, or 2

4th root k = 0, 1, 2, or 3

etc.