1) Find all the roots of: (-2 + 2i) 1/3
We use: wn = 3Ör [cos(q + 2 pk/n) + isin[(q + 2 pk/n)]
Note: 3Ör= r 1/3 = [Ö8]1/3
BUT: [Ö8]1/3 = Ö2
The roots are then defined:
Note: 3Ör= r 1/3 = [Ö8]1/3
BUT: [Ö8]1/3 = Ö2
The roots are then defined:
wn = Ö2 [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 = Ö2 [cos(p/4 + 2 pk/3) + isin(p/4 + 2 pk/3)]
So the consecutive roots are(in order):
w0 = Ö2 [cos(p/4 + 2 pk/3) + isin(p/4 + 2 pk/3)]
since k=0 for w0:
w0 = Ö2 [cos(p/4) + isin(p/4)] = 1 +i
Since k = 1 for w1
w1 = Ö2 [cos(p/4 + 2 p/3) + isin(p/4 + 2 p/3)]
Working on the interior brackets (adding fractions in p):
w1 = Ö2 [cos(p/4 + 2 p/3) + isin(p/4 + 2 p/3)]
Working on the interior brackets (adding fractions in p):
(p/4 + 2 p/3) = (3 p + 8p)/ 12 = 11 p/12
So: w1 = Ö2 [cos(11 p / 12) + isin(11 p/12)]
Finally since k =2 (for w2):
w2 = Ö2 [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 = Ö2 [cos(19 p/12 ) + isin(19 p/12)]
Let’s check one of the roots: (1 +i):
(1 +i) (1 +i)( 1 +i) = (-2 + 2i) ?
Take: (1 +i) (1 +i) = 1 +2i -1 = 2i
Then:
2i((1 +i) = 2i + 2i(i) = 2i -2 so that (1 + i) is a third root of (-2 + 2i)!
2) Find all the roots of (-16)1/4
Arg(z) = p/4 and r = 16
The summarized (compacted, i.e. 'cis') form for the computable root formula to find all roots is:
wn = 4Ö(16 cis p) = 2 cis [p/4, 3p/4, 5p/4, 7p/4]
for k = 0, 1, 2, 3 , which yields:
wo = 2 [cos(p/4) + isin(p/4)] = Ö2 (1 + i)
w1 = 2 [cos(3p/4) + isin(3p/4)] = Ö2 (-1 + i)
w2 = 2 [cos(5p/4) + isin(5p/4)] = Ö2 (-1 - i)
w3 = 2 [cos(7p/4) + isin(7p/4)] = Ö2 (1 - i)
Check one of the roots: [Ö2 (1 + i)] 4 = -16 (Checked using MathCad)
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