## Thursday, March 14, 2013

### More Complex Roots

What about complex roots not of unity?

Say: (1 + i)1/3

For this example, we use:

wn3Ö [cos(q + 2 pk/n) + isin[(q + 2 pk/n)]

Note: 3Ör=  r 1/3 = [Ö2]1/3(2)1/ 6

The roots are then defined from:

wn = (2)1/ 6 [cos(q + 2 pk/3) + isin(q + 2 pk/3)]

Since q = arctan y/x = 45 deg then, q = p/4

So the consecutive roots are(in order):

w0 = (2)1/6 [cos(p/4 + 2 pk/3) + isin(p/4 + 2 pk/3)]

Since k=0:

w0 = (2)1/6 [cos(p/4) + isin(p/4)]

Further:

w1 = (2)1/6 [cos(p/4 + 2 pk/3)  + isin(p/4 + 2 pk/3)]

Working on the interior brackets (adding fractions in p):

(p/4 + 2 p/3)  = (3 p + 8p)/ 12  = 11 p/12

So:

w1 = (2)1/6 [cos(11 p / 12) + isin(11 p/12)]

and the answer can be left in this form, OR the reader can obtain cos(11p/12) and sin(11p/12) and work out the specific numbers.

Finally since k =2 (for w2):

w2 = (2)1/6 [cos(p/4 + 4p/3) + isin(p/4 + 4p/3)]

Again, adding the fractions in p in the interior brackets:

(p/4 + 4p/3)  = (3 p + 16 p)/ 12 = 19 p/12

So:

w2 = (2)1/6 [cos(19 p/12 ) + isin(19 p/12)]

Problems for the Math Maven:

1)  Find all the roots of: (-2 + 2i) 1/3
2) Find all the roots of (-16)1/4