Solutions to Complex Roots Problems

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

We use: w_n = 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:

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):

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):

(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:

w_n = ⁴Ö(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)

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)