1) Find all the roots of: (-2 + 2i)

^{1/3}
We use: w

Note:

BUT: [Ö8]

The roots are then defined:

_{n}=^{3}Ör [cos(q + 2 pk/n) + isin[(q + 2 pk/n)]Note:

^{3}Ör= r^{1/3}= [Ö8]^{1/3}BUT: [Ö8]

^{1/3}= Ö2The roots are then defined:

w

_{n}= 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):

w

So the consecutive roots are(in order):

w

_{0}= Ö2 [cos(p/4 + 2 pk/3) + isin(p/4 + 2 pk/3)]
since k=0 for w

_{0}:
w

_{0 }= Ö2 [cos(p/4) + isin(p/4)] = 1 +i
Since k = 1 for w

w

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

_{1 }w

_{1}= 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: w

_{1 }= Ö2 [cos(11 p / 12) + isin(11 p/12)]

Finally since k =2 (for w

_{2}):

w

_{2}= 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: w

_{2 }= 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:

w

_{n}=^{4}Ö(16 cis p) = 2 cis [p/4, 3p/4, 5p/4, 7p/4]for k = 0, 1, 2, 3 , which yields:

w

_{o}= 2 [cos(p/4) + isin(p/4)] = Ö2 (1 + i)
w

_{1}= 2 [cos(3p/4) + isin(3p/4)] = Ö2 (-1 + i)
w

_{2}= 2 [cos(5p/4) + isin(5p/4)] = Ö2 (-1 - i)
w

_{3}= 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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