Say: (1 + i)1/3
For this example, we use:
wn = 3Ör [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)]
wn = 3Ör [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.
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
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
No comments:
Post a Comment